Difference between revisions of "Integration by Parts"

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::<math>\begin{array}{rcl}
 
::<math>\begin{array}{rcl}
−
\displaystyle{f(x)g(x)} & = & \displaystyle{\int f'(x)g(x)+f(x)g'(x)~dx}\\
+
\displaystyle{f(x)g(x)} & = & \displaystyle{\int (f(x)g(x))'~dx}\\
 +
&&\\
 +
& = & \displaystyle{\int f'(x)g(x)+f(x)g'(x)~dx}\\
 
&&\\
 
&&\\
 
& = & \displaystyle{\int f'(x)g(x)~dx+\int f(x)g'(x)~dx.}
 
& = & \displaystyle{\int f'(x)g(x)~dx+\int f(x)g'(x)~dx.}
Line 51: Line 53:
 
Evaluate the following integrals.
 
Evaluate the following integrals.
  
−
'''1)''' &nbsp; <math style="vertical-align: -7px">\int xe^x~dx</math>
+
'''1)''' &nbsp; <math style="vertical-align: -13px">\int xe^x~dx</math>
  
 
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Solution: &nbsp;
 
!Solution: &nbsp;
 
|-
 
|-
−
|Using the Product Rule, we have  
+
|We have two options when doing integration by parts.  
−
|-
 
−
|
 
−
::<math>f'(x)=(x^2+x+1)(x^3+2x^2+4)'+(x^2+x+1)'(x^3+2x^2+4).</math>
 
−
|-
 
−
|Then, using the Power Rule, we have
 
 
|-
 
|-
−
|
+
|We can let &nbsp;<math style="vertical-align: 0px">u=x</math>&nbsp; or &nbsp;<math style="vertical-align: 0px">u=e^x.</math>
−
::<math>f'(x)=(x^2+x+1)(3x^2+4x)+(2x+1)(x^3+2x^2+4).</math>
 
 
|-
 
|-
 +
|In this case, we let &nbsp;<math style="vertical-align: 0px">u</math>&nbsp; be the polynomial.
 
|-
 
|-
−
|<u>NOTE:</u> It is not necessary to use the Product Rule to calculate the derivative of this function.
+
|So, we let &nbsp;<math style="vertical-align: 0px">u=x</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=e^x~dx.</math>
 
|-
 
|-
−
|You can distribute the terms and then use the Power Rule.
+
|Then, &nbsp;<math style="vertical-align: 0px">du=dx</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">v=e^x.</math>
 
|-
 
|-
−
|In this case, we have
+
|Hence, by integration by parts, we get
 
|-
 
|-
 
|
 
|
 
::<math>\begin{array}{rcl}
 
::<math>\begin{array}{rcl}
−
\displaystyle{f(x)} & = & \displaystyle{(x^2+x+1)(x^3+2x^2+4)}\\
+
\displaystyle{\int xe^x~dx} & = & \displaystyle{xe^x-\int e^x~dx}\\
−
&&\\
 
−
& = & \displaystyle{x^2(x^3+2x^2+4)+x(x^3+2x^2+4)+1(x^3+2x^2+4)}\\
 
−
&&\\
 
−
& = & \displaystyle{x^5+2x^4+4x^2+x^4+2x^3+4x+x^3+2x^2+4} \\
 
 
&&\\
 
&&\\
−
& = & \displaystyle{x^5+3x^4+3x^3+6x^2+4x+4.}
+
& = & \displaystyle{xe^x-e^x+C.}
 
\end{array}</math>
 
\end{array}</math>
−
|-
 
−
|Now, using the Power Rule, we get
 
−
|-
 
−
|
 
−
::<math>f'(x)=5x^4+12x^3+9x^2+12x+4.</math>
 
−
|-
 
−
|In general, calculating derivatives in this way is tedious. It would be better to use the Product Rule.
 
 
|}
 
|}
  
Line 95: Line 81:
 
!Final Answer: &nbsp;
 
!Final Answer: &nbsp;
 
|-
 
|-
−
|&nbsp; &nbsp; &nbsp; &nbsp;<math>f'(x)=(x^2+x+1)(3x^2+4x)+(2x+1)(x^3+2x^2+4)</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math>xe^x-e^x+C</math>
−
|-
 
−
|or equivalently
 
−
|-
 
−
|&nbsp; &nbsp; &nbsp; &nbsp;<math>f'(x)=x^5+3x^4+3x^3+6x^2+4x+4</math>
 
 
|}
 
|}
  
Line 108: Line 90:
 
|-
 
|-
 
|
 
|
−
Using the Quotient Rule, we have
+
We have a choice to make.
 
|-
 
|-
−
|
+
|We can let &nbsp;<math style="vertical-align: 0px">u=x</math>&nbsp; or &nbsp;<math style="vertical-align: -5px">u=\cos(2x).</math>
−
::<math>f'(x)=\frac{x(x^2+x^3)'-(x^2+x^3)(x)'}{x^2}.</math>
+
|-
 +
|In this case, we let &nbsp;<math style="vertical-align: 0px">u</math>&nbsp; be the polynomial.
 
|-
 
|-
−
|Then, using the Power Rule, we have
+
|So, we let &nbsp;<math style="vertical-align: 0px">u=x</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">dv=\cos(2x)~dx.</math>
 
|-
 
|-
−
|
+
|Then, &nbsp;<math style="vertical-align: 0px">du=dx.</math>&nbsp;
−
::<math>f'(x)=\frac{x(2x+3x^2)-(x^2+x^3)(1)}{x^2}.</math>
 
 
|-
 
|-
−
|<u>NOTE:</u> It is not necessary to use the Quotient Rule to calculate the derivative of this function.
+
|To find &nbsp;<math style="vertical-align: -4px">v,</math>&nbsp; we need to use &nbsp;<math style="vertical-align: 0px">u-</math>substitution. So,
 
|-
 
|-
−
|You can divide and then use the Power Rule.
+
|
 +
::<math style="vertical-align: -13px">v=\int \cos(2x)~dx=\frac{1}{2} \sin(2x).</math>
 
|-
 
|-
−
|In this case, we have
+
|Hence, by integration by parts, we get
 
|-
 
|-
 
|
 
|
 
::<math>\begin{array}{rcl}
 
::<math>\begin{array}{rcl}
−
\displaystyle{f(x)} & = & \displaystyle{\frac{x^2+x^3}{x}}\\
+
\displaystyle{\int x\cos(2x)~dx} & = & \displaystyle{\frac{1}{2}x\sin(2x)-\int \frac{1}{2}\sin(2x)~dx}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{\frac{x^2}{x}+\frac{x^3}{x}}\\
+
& = & \displaystyle{\frac{1}{2}x\sin(2x)+\frac{1}{4}\cos(2x)+C,}
−
&&\\
 
−
& = & \displaystyle{x+x^2.} \\
 
 
\end{array}</math>
 
\end{array}</math>
 
|-
 
|-
−
|Now, using the Power Rule, we get
+
|where we use &nbsp;<math style="vertical-align: 0px">u-</math> substitution to evaluate the last integral.
−
|-
 
−
|
 
−
::<math>f'(x)=1+2x.</math>
 
 
|}
 
|}
  
Line 142: Line 120:
 
!Final Answer: &nbsp;
 
!Final Answer: &nbsp;
 
|-
 
|-
−
||&nbsp; &nbsp; &nbsp; &nbsp;<math>f'(x)=\frac{x(2x+3x^2)-(x^2+x^3)}{x^2}</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{1}{2}x\sin(2x)+\frac{1}{4}\cos(2x)+C</math>
−
|-
 
−
|or equivalently
 
−
|-
 
−
|&nbsp; &nbsp; &nbsp; &nbsp;<math>f'(x)=1+2x</math>
 
−
 
 
−
|-
 
 
|}
 
|}
  
Line 156: Line 128:
 
!Solution: &nbsp;
 
!Solution: &nbsp;
 
|-
 
|-
−
|Using the Quotient Rule, we get
+
|We have a choice to make.
 +
|-
 +
|We can let &nbsp;<math style="vertical-align: -1px">u=\ln x</math>&nbsp; or &nbsp;<math style="vertical-align: -1px">dv=\ln x~dx.</math>
 +
|-
 +
|In this case, we don't want to let &nbsp;<math style="vertical-align: -1px">dv=\ln x~dx</math>&nbsp; since we don't know how to integrate &nbsp;<math style="vertical-align: -1px">\ln x</math>&nbsp; yet.
 +
|-
 +
|So, we let &nbsp;<math style="vertical-align: -1px">u=\ln x</math>&nbsp; and &nbsp;<math style="vertical-align: -1px">dv=1~dx.</math>
 +
|-
 +
|Then, &nbsp;<math style="vertical-align: -13px">du=\frac{1}{x}~dx</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">v=x.</math>
 +
|-
 +
|Hence, by integration by parts, we get  
 
|-
 
|-
 
|
 
|
 
::<math>\begin{array}{rcl}
 
::<math>\begin{array}{rcl}
−
\displaystyle{f'(x)} & = & \displaystyle{\frac{\cos x(\sin x)'-\sin x (\cos x)'}{(\cos x)^2}}\\
+
\displaystyle{\int \ln x~dx} & = & \displaystyle{x\ln (x) -\int x \bigg(\frac{1}{x}\bigg)~dx}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{\frac{\cos x(\cos x)-\sin x (-\sin x)}{(\cos x)^2}}\\
+
& = & \displaystyle{x\ln (x)-\int 1~dx}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{\frac{\cos^2 x+\sin^2 x}{\cos^2 x}} \\
+
& = & \displaystyle{x\ln (x)-x+C.}
−
&&\\
 
−
& = & \displaystyle{\frac{1}{\cos^2 x}}\\
 
−
&&\\
 
−
& = & \displaystyle{\sec^2 x}
 
 
\end{array}</math>
 
\end{array}</math>
 
|-
 
|-
−
|since &nbsp;<math style="vertical-align: -2px">\sin^2 x+\cos^2 x=1</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">\sec x=\frac{1}{\cos x}.</math>
+
|<u>Note:</u> The domain of the function &nbsp;<math style="vertical-align: -1px">\ln x</math>&nbsp; is &nbsp;<math style="vertical-align: -5px">(0,\infty).</math>
 
|-
 
|-
−
|Since &nbsp;<math style="vertical-align: -14px">\frac{\sin x}{\cos x}=\tan x,</math>&nbsp; we have
+
|So, this integral is defined for &nbsp;<math style="vertical-align: 0px">x>0.</math>
−
|-
 
−
|
 
−
::<math>\frac{d}{dx}{\tan x}=\sec^2 x.</math>
 
 
|}
 
|}
  
Line 182: Line 157:
 
!Final Answer: &nbsp;
 
!Final Answer: &nbsp;
 
|-
 
|-
−
|&nbsp; &nbsp; &nbsp; &nbsp;<math>f'(x)=\sec^2 x</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math>x\ln (x)-x+C</math>
 
|-
 
|-
 
|}
 
|}
Line 188: Line 163:
 
== Exercise 1 ==
 
== Exercise 1 ==
  
−
Evaluate &nbsp;<math style="vertical-align: -13px">\int x^3 e^{-2x}~dx.</math>
+
Evaluate &nbsp;<math style="vertical-align: -13px">\int x \sec^2 x~dx.</math>
 +
 
 +
Since we know the antiderivative of &nbsp;<math style="vertical-align: -3px">\sec^2 x,</math>
  
−
First, we need to know the derivative of &nbsp;<math style="vertical-align: 0px">\csc x.</math>&nbsp; Recall
+
we let &nbsp;<math style="vertical-align: 0px">u=x</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=\sec^2 x~dx.</math>
  
−
::<math>\csc x =\frac{1}{\sin x}.</math>
+
Then, &nbsp;<math style="vertical-align: 0px">du=dx</math>&nbsp; and &nbsp;<math style="vertical-align: -1px">v=\tan x.</math>
  
−
Now, using the Quotient Rule, we have
+
Using integration by parts, we get
  
 
::<math>\begin{array}{rcl}
 
::<math>\begin{array}{rcl}
−
\displaystyle{\frac{d}{dx}(\csc x)} & = & \displaystyle{\frac{d}{dx}\bigg(\frac{1}{\sin x}\bigg)}\\
+
\displaystyle{\int x \sec^2 x~dx} & = & \displaystyle{x\tan x -\int \tan x~dx}\\
−
&&\\
 
−
& = & \displaystyle{\frac{\sin x (1)'-1(\sin x)'}{\sin^2 x}}\\
 
−
&&\\
 
−
& = & \displaystyle{\frac{\sin x (0)-\cos x}{\sin^2 x}}\\
 
−
&&\\
 
−
& = & \displaystyle{\frac{-\cos x}{\sin^2 x}} \\
 
 
&&\\
 
&&\\
−
& = & \displaystyle{-\csc x \cot x.}
+
& = & \displaystyle{x\tan x -\int \frac{\sin x}{\cos x}~dx.}
 
\end{array}</math>
 
\end{array}</math>
  
−
Using the Product Rule and Power Rule, we have
+
For the remaining integral, we use &nbsp;<math style="vertical-align: 0px">u-</math>substitution.
 +
 
 +
Let &nbsp;<math style="vertical-align: 0px">u=\cos x.</math>&nbsp; Then, &nbsp;<math style="vertical-align: -1px">du=-\sin x~dx.</math>
 +
 
 +
So, we get
  
 
::<math>\begin{array}{rcl}
 
::<math>\begin{array}{rcl}
−
\displaystyle{f'(x)} & = & \displaystyle{\frac{1}{x^2}(\csc x-4)'+\bigg(\frac{1}{x^2}\bigg)'(\csc x-4)}\\
+
\displaystyle{\int x \sec^2 x~dx} & = & \displaystyle{x\tan x +\int \frac{1}{u}~du}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{\frac{1}{x^2}(-\csc x \cot x+0)+(-2x^{-3})(\csc x-4)}\\
+
& = & \displaystyle{x\tan x + \ln |u|+C}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{\frac{-\csc x \cot x}{x^2}+\frac{-2(\csc x-4)}{x^3}.}
+
& = & \displaystyle{x\tan x+ \ln |\cos x|+C.}
 
\end{array}</math>
 
\end{array}</math>
  
 
So, we have  
 
So, we have  
−
::<math>f'(x)=\frac{-\csc x \cot x}{x^2}+\frac{-2(\csc x-4)}{x^3}.</math>
+
::<math>\int x \sec^2 x~dx=x\tan x+ \ln |\cos x|+C.</math>
  
 
== Exercise 2 ==
 
== Exercise 2 ==
  
−
Evaluate &nbsp;<math style="vertical-align: -5px">\int e^{3x}\sin (2x)~dx.</math>
+
Evaluate &nbsp;<math style="vertical-align: -13px">\int \frac{\ln x}{x^3}~dx.</math>
  
−
Notice that the function &nbsp;<math style="vertical-align: -5px">g(x)</math>&nbsp; is the product of three functions.  
+
We start by letting &nbsp;<math style="vertical-align: -1px">u=\ln x</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">dv=\frac{1}{x^3}~dx.</math>
  
−
We start by grouping two of the functions together. So, we have &nbsp;<math style="vertical-align: -5px">g(x)=(2x\sin x)\sec x.</math>
+
Then, &nbsp;<math style="vertical-align: -13px">du=\frac{1}{x}~dx</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">v=-\frac{1}{2x^2}.</math>
  
−
Using the Product Rule, we get
+
So, using integration by parts, we get
  
 
::<math>\begin{array}{rcl}
 
::<math>\begin{array}{rcl}
−
\displaystyle{g'(x)} & = & \displaystyle{(2x\sin x)(\sec x)'+(2x\sin x)'\sec x}\\
+
\displaystyle{\int  \frac{\ln x}{x^3}~dx} & = & \displaystyle{-\frac{\ln x}{2x^2}-\int -\frac{1}{2x^2}\bigg(\frac{1}{x}\bigg)~dx}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{(2x\sin x)(\tan^2 x)+(2x\sin x)'\sec x.}
+
& = & \displaystyle{-\frac{\ln x}{2x^2}+\int \frac{1}{2x^3}~dx}\\
−
\end{array}</math>
 
−
 
 
−
Now, we need to use the Product Rule again. So,
 
−
 
 
−
::<math>\begin{array}{rcl}
 
−
\displaystyle{g'(x)} & = & \displaystyle{2x\sin x\tan^2 x+(2x(\sin x)'+(2x)'\sin x)\sec x}\\
 
 
&&\\
 
&&\\
−
& = & \displaystyle{2x\sin x\tan^2 x+(2x\cos x+2\sin x)\sec x.}
+
& = & \displaystyle{-\frac{\ln x}{2x^2}-\frac{1}{4x^2}+C.}
 
\end{array}</math>
 
\end{array}</math>
  
 
So, we have  
 
So, we have  
−
::<math>g'(x)=2x\sin x\tan^2 x+(2x\cos x+2\sin x)\sec x.</math>
 
  
−
But, there is another way to do this problem. Notice
+
::<math>\int \frac{\ln x}{x^3}~dx=-\frac{\ln x}{2x^2}-\frac{1}{4x^2}+C.</math>
  
−
::<math>\begin{array}{rcl}
+
== Exercise 3 ==
−
\displaystyle{g(x)} & = & \displaystyle{2x\sin x\sec x}\\
+
Evaluate &nbsp;<math style="vertical-align: -13px">\int x\sqrt{x+1}~dx.</math>
−
&&\\
 
−
& = & \displaystyle{2x\sin x\frac{1}{\cos x}}\\
 
−
&&\\
 
−
& = & \displaystyle{2x\tan x.}
 
−
\end{array}</math>
 
  
−
Now, you would only need to use the Product Rule once instead of twice.
+
We start by letting &nbsp;<math style="vertical-align: 0px">u=x</math>&nbsp; and &nbsp;<math style="vertical-align: -3px">dv=\sqrt{x+1}~dx.</math>
−
 
 
−
== Exercise 3 ==
 
  
−
Evaluate &nbsp;<math style="vertical-align: -16px">\int x\sec^2 x~dx.</math>
+
Then, &nbsp;<math style="vertical-align: 0px">du=dx.</math>
  
−
Using the Quotient Rule, we have
+
To find &nbsp;<math style="vertical-align: -4px">v,</math>&nbsp; we need to use &nbsp;<math style="vertical-align: 0px">u-</math>substitution. So,
  
−
::<math>h'(x)=\frac{(x^2\cos x+3)(x^2\sin x+1)'-(x^2\sin x+1)(x^2\cos x+3)'}{(x^2\cos x+3)^2}.</math>
+
::<math style="vertical-align: 0px">v=\int \sqrt{x+1}~dx=\frac{2}{3}(x+1)^{\frac{3}{2}}.</math>
  
−
Now, we need to use the Product Rule. So, we have
+
Hence, by integration by parts, we get
  
 
::<math>\begin{array}{rcl}
 
::<math>\begin{array}{rcl}
−
\displaystyle{h'(x)} & = & \displaystyle{\frac{(x^2\cos x+3)(x^2(\sin x)'+(x^2)'\sin x)-(x^2\sin x+1)(x^2(\cos x)'+(x^2)'\cos x)}{(x^2\cos x+3)^2}}\\
+
\displaystyle{\int x\sqrt{x+1}~dx} & = & \displaystyle{\frac{2}{3}x(x+1)^{\frac{3}{2}}-\int \frac{2}{3}(x+1)^{\frac{3}{2}}~dx}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{\frac{(x^2\cos x+3)(x^2\cos x+2x\sin x)-(x^2\sin x+1)(-x^2\sin x+2x\cos x)}{(x^2\cos x+3)^2}.}
+
& = & \displaystyle{\frac{2}{3}x(x+1)^{\frac{3}{2}}-\frac{4}{15}(x+1)^{\frac{5}{2}}+C,}
 
\end{array}</math>
 
\end{array}</math>
  
−
So, we get
+
where we use &nbsp;<math style="vertical-align: 0px">u-</math> substitution to evaluate the last integral.
−
::<math>h'(x)=\frac{(x^2\cos x+3)(x^2\cos x+2x\sin x)-(x^2\sin x+1)(-x^2\sin x+2x\cos x)}{(x^2\cos x+3)^2}.</math>
+
 
 +
So, we have
 +
 
 +
::<math>\int x\sqrt{x+1}~dx=\frac{2}{3}x(x+1)^{\frac{3}{2}}-\frac{4}{15}(x+1)^{\frac{5}{2}}+C.</math>
  
 
== Exercise 4 ==
 
== Exercise 4 ==
  
−
Evaluate  &nbsp;<math style="vertical-align: -14px">\int x\sqrt{x+1}~dx.</math>
+
Evaluate  &nbsp;<math style="vertical-align: -13px">\int x^2e^{-2x}~dx.</math>
 +
 
 +
We start by letting &nbsp;<math style="vertical-align: 0px">u=x^2</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=e^{-2x}~dx.</math>
  
−
First, using the Quotient Rule, we have
+
Then, &nbsp;<math style="vertical-align: 0px">du=2x~dx.</math>
 +
 
 +
To find &nbsp;<math style="vertical-align: -4px">v,</math>&nbsp; we need to use &nbsp;<math style="vertical-align: 0px">u-</math>substitution. So,
 +
 
 +
::<math style="vertical-align: 0px">v=\int e^{-2x}~dx=\frac{e^{-2x}}{-2}.</math>
 +
 
 +
Hence, by integration by parts, we get
  
 
::<math>\begin{array}{rcl}
 
::<math>\begin{array}{rcl}
−
\displaystyle{f'(x)} & = & \displaystyle{\frac{x^2\sin x (e^x)'-e^x(x^2\sin x)'}{(x^2\sin x)^2}}\\
+
\displaystyle{\int x^2 e^{-2x}~dx} & = & \displaystyle{\frac{x^2e^{-2x}}{-2}-\int 2x\bigg(\frac{e^{-2x}}{-2}\bigg)~dx}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2\sin x)'}{x^4\sin^2 x}.}
+
& = & \displaystyle{\frac{x^2e^{-2x}}{-2}+\int xe^{-2x}~dx.}
 
\end{array}</math>
 
\end{array}</math>
  
−
Now, we need to use the Product Rule. So, we have
+
Now, we need to use integration by parts a second time.
 +
 
 +
Let &nbsp;<math style="vertical-align: 0px">u=x</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=e^{-2x}~dx.</math>
 +
 
 +
Then, &nbsp;<math style="vertical-align: 0px">du=dx</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">v=\frac{e^{-2x}}{-2}.</math>
 +
 
 +
Therefore, using integration by parts again, we get
  
 
::<math>\begin{array}{rcl}
 
::<math>\begin{array}{rcl}
−
\displaystyle{f'(x)} & = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2(\sin x)'+(x^2)'\sin x)}{x^4\sin^2 x}}\\
+
\displaystyle{\int x^2 e^{-2x}~dx} & = & \displaystyle{\frac{x^2 e^{-2x} }{-2}+\frac{x e^{-2x} }{-2}-\int \frac{e^{-2x}}{-2}~dx}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2\cos x+2x\sin x)}{x^4\sin^2 x}.}
+
& = & \displaystyle{\frac{x^2e^{-2x}}{-2}+\frac{xe^{-2x}}{-2}+\frac{e^{-2x}}{-4}+C.}
 
\end{array}</math>
 
\end{array}</math>
  
 
So, we have  
 
So, we have  
−
::<math>f'(x)=\frac{x^2\sin x e^x - e^x(x^2\cos x+2x\sin x)}{x^4\sin^2 x}.</math>
+
::<math>\int x^2e^{-2x}~dx=\frac{x^2e^{-2x}}{-2}+\frac{xe^{-2x}}{-2}+\frac{e^{-2x}}{-4}+C.</math>
  
 
== Exercise 5 ==
 
== Exercise 5 ==
  
−
Evaluate  &nbsp;<math style="vertical-align: -14px">\int \frac{\ln x}{x^3}~dx.</math>
+
Evaluate  &nbsp;<math style="vertical-align: -13px">\int e^{3x}\sin(2x)~dx.</math>
  
−
First, using the Quotient Rule, we have
+
We begin by letting &nbsp;<math style="vertical-align: -6px">u=\sin(2x)</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=e^{3x}~dx.</math>
 +
 
 +
Then, &nbsp;<math style="vertical-align: -5px">du=2\cos (2x)~dx.</math>
 +
 
 +
To find &nbsp;<math style="vertical-align: -4px">v,</math>&nbsp; we need to use &nbsp;<math style="vertical-align: 0px">u-</math>substitution. So,
 +
 
 +
::<math style="vertical-align: 0px">v=\int e^{3x}~dx=\frac{e^{3x}}{3}.</math>
 +
 
 +
Hence, by integration by parts, we have
  
 
::<math>\begin{array}{rcl}
 
::<math>\begin{array}{rcl}
−
\displaystyle{f'(x)} & = & \displaystyle{\frac{x^2\sin x (e^x)'-e^x(x^2\sin x)'}{(x^2\sin x)^2}}\\
+
\displaystyle{\int e^{3x} \sin(2x)~dx} & = & \displaystyle{\frac{e^{3x}\sin(2x)}{3}-\int \frac{2}{3}\cos(2x)e^{3x}~dx}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2\sin x)'}{x^4\sin^2 x}.}
+
& = & \displaystyle{\frac{e^{3x}\sin(2x)}{3}-\frac{2}{3} \int \cos(2x)e^{3x}~dx.}
 
\end{array}</math>
 
\end{array}</math>
  
−
Now, we need to use the Product Rule. So, we have
+
Now, we need to use integration by parts a second time.
 +
 
 +
Let &nbsp;<math style="vertical-align: -5px">u=\cos (2x)</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=e^{3x}~dx.</math>
 +
 
 +
Then, &nbsp;<math style="vertical-align: -5px">du=-2\sin(2x)~dx</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">v=\frac{e^{3x}}{3}.</math>
 +
 
 +
Therefore, using integration by parts again, we get
  
 
::<math>\begin{array}{rcl}
 
::<math>\begin{array}{rcl}
−
\displaystyle{f'(x)} & = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2(\sin x)'+(x^2)'\sin x)}{x^4\sin^2 x}}\\
+
\displaystyle{\int e^{3x} \sin(2x)~dx} & = & \displaystyle{\frac{e^{3x}\sin(2x)}{3}-\frac{2}{3} \bigg[ \frac{\cos(2x)e^{3x}}{3}+\int \frac{2}{3}\sin(2x)e^{3x}~dx\bigg]}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2\cos x+2x\sin x)}{x^4\sin^2 x}.}
+
& = & \displaystyle{\frac{e^{3x}\sin(2x)}{3}-\frac{2\cos(2x)e^{3x}}{9}-\frac{4}{9}\int e^{3x}\sin(2x)~dx.}
 
\end{array}</math>
 
\end{array}</math>
  
−
So, we have
+
Now, we have the exact same integral that we had at the beginning of the problem.
−
::<math>f'(x)=\frac{x^2\sin x e^x - e^x(x^2\cos x+2x\sin x)}{x^4\sin^2 x}.</math>
+
 
 +
So, we add this integral to the other side of the equation.
 +
 
 +
When we do this, we get
 +
 
 +
::<math>\frac{13}{9} \int e^{3x}\sin(2x)~dx = \frac{e^{3x}\sin(2x)}{3}-\frac{2\cos(2x)e^{3x}}{9}.</math>
 +
 
 +
Therefore, we get
 +
 
 +
::<math> \int e^{3x}\sin(2x)~dx = \frac{9}{13}\bigg(\frac{e^{3x}\sin(2x)}{3}-\frac{2\cos(2x)e^{3x}}{9}\bigg)+C.</math>
  
 
== Exercise 6 ==
 
== Exercise 6 ==
Line 329: Line 330:
 
Evaluate  &nbsp;<math style="vertical-align: -14px">\int \sin(2x)\cos(3x)~dx.</math>
 
Evaluate  &nbsp;<math style="vertical-align: -14px">\int \sin(2x)\cos(3x)~dx.</math>
  
−
First, using the Quotient Rule, we have
+
For this problem, we use a similar process as Exercise 5.
 +
 
 +
We use integration by parts twice, which produces the same integral given to us in the problem.
 +
 
 +
Then, we solve for our integral.
 +
 
 +
We begin by letting &nbsp;<math style="vertical-align: -5px">u=\sin(2x)</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">dv=\cos(3x)~dx.</math>
 +
 
 +
Then, &nbsp;<math style="vertical-align: -5px">du=2\cos (2x)~dx.</math>
 +
 
 +
To find &nbsp;<math style="vertical-align: -4px">v,</math>&nbsp; we need to use &nbsp;<math style="vertical-align: 0px">u-</math>substitution. So,
 +
 
 +
::<math style="vertical-align: 0px">v=\int \cos(3x)~dx=\frac{1}{3}\sin(3x).</math>
 +
 
 +
Hence, by integration by parts, we have
  
 
::<math>\begin{array}{rcl}
 
::<math>\begin{array}{rcl}
−
\displaystyle{f'(x)} & = & \displaystyle{\frac{x^2\sin x (e^x)'-e^x(x^2\sin x)'}{(x^2\sin x)^2}}\\
+
\displaystyle{\int \sin(2x)\cos(3x)~dx} & = & \displaystyle{\frac{1}{3}\sin(2x)\sin(3x)-\int \frac{2}{3}\cos(2x)\sin(3x)~dx}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2\sin x)'}{x^4\sin^2 x}.}
+
& = & \displaystyle{\frac{1}{3}\sin(2x)\sin(3x)-\frac{2}{3} \int \cos(2x)\sin(3x)~dx.}
 
\end{array}</math>
 
\end{array}</math>
  
−
Now, we need to use the Product Rule. So, we have
+
Now, we need to use integration by parts a second time.
 +
 
 +
Let &nbsp;<math style="vertical-align: -5px">u=\cos (2x)</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">dv=\sin(3x)~dx.</math>
 +
 
 +
Then, &nbsp;<math style="vertical-align: -5px">du=-2\sin(2x)~dx</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">v=\frac{-\cos(3x)}{3}.</math>
 +
 
 +
Therefore, using integration by parts again, we get
  
 
::<math>\begin{array}{rcl}
 
::<math>\begin{array}{rcl}
−
\displaystyle{f'(x)} & = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2(\sin x)'+(x^2)'\sin x)}{x^4\sin^2 x}}\\
+
\displaystyle{\int \sin(2x)\cos(3x)~dx} & = & \displaystyle{\frac{1}{3}\sin(2x)\sin(3x)-\frac{2}{3} \bigg[ \frac{-\cos(2x)\cos(3x)}{3}-\int \frac{2}{3}\sin(2x)\cos(3x)~dx\bigg]}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2\cos x+2x\sin x)}{x^4\sin^2 x}.}
+
& = & \displaystyle{\frac{1}{3}\sin(2x)\sin(3x)+ \frac{2\cos(2x)\cos(3x)}{9}+\frac{4}{9}\int \sin(2x)\cos(3x)~dx.}
 
\end{array}</math>
 
\end{array}</math>
  
−
So, we have
+
Now, we have the exact same integral that we had at the beginning of the problem.
−
::<math>f'(x)=\frac{x^2\sin x e^x - e^x(x^2\cos x+2x\sin x)}{x^4\sin^2 x}.</math>
+
 
 +
So, we subtract this integral to the other side of the equation.
 +
 
 +
When we do this, we get
 +
 
 +
::<math>\frac{5}{9} \int \sin(2x)\cos(3x)~dx = \frac{1}{3}\sin(2x)\sin(3x)+ \frac{2\cos(2x)\cos(3x)}{9}.</math>
 +
 
 +
Therefore, we get
 +
 
 +
::<math> \int e^{3x}\sin(2x)~dx = \frac{9}{5}\bigg(\frac{1}{3}\sin(2x)\sin(3x)+ \frac{2\cos(2x)\cos(3x)}{9}\bigg)+C.</math>

Latest revision as of 12:17, 30 October 2017

Introduction

Let's say we want to integrate

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int x^{2}e^{x^{3}}~dx.}

Here, we can compute this antiderivative by using  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u-} substitution.

While  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u-} substitution is an important integration technique, it will not help us evaluate all integrals.

For example, consider the integral

There is no substitution that will allow us to integrate this integral.

We need another integration technique called integration by parts.

The formula for integration by parts comes from the product rule for derivatives.

Recall from the product rule,

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (f(x)g(x))'=f'(x)g(x)+f(x)g'(x).}

Then, we have

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {f(x)g(x)}&=&\displaystyle {\int (f(x)g(x))'~dx}\\&&\\&=&\displaystyle {\int f'(x)g(x)+f(x)g'(x)~dx}\\&&\\&=&\displaystyle {\int f'(x)g(x)~dx+\int f(x)g'(x)~dx.}\end{array}}}

If we solve the last equation for the second integral, we obtain

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int f(x)g'(x)~dx=f(x)g(x)-\int f'(x)g(x)~dx.}

This formula is the formula for integration by parts.

But, as it is currently stated, it is long and hard to remember.

So, we make a substitution to obtain a nicer formula.

Let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=f(x)}   and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle dv=g'(x)~dx.}

Then,    and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle v=g(x).}

Plugging these into our formula, we obtain

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int u~dv=uv-\int v~du.}

Warm-Up

Evaluate the following integrals.

1)   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int xe^{x}~dx}

Solution:  
We have two options when doing integration by parts.
We can let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=x}   or  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=e^{x}.}
In this case, we let    be the polynomial.
So, we let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=x}   and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle dv=e^{x}~dx.}
Then,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle du=dx}   and  
Hence, by integration by parts, we get
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {\int xe^{x}~dx}&=&\displaystyle {xe^{x}-\int e^{x}~dx}\\&&\\&=&\displaystyle {xe^{x}-e^{x}+C.}\end{array}}}
Final Answer:  
       Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle xe^{x}-e^{x}+C}

2)   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int x\cos(2x)~dx}

Solution:  

We have a choice to make.

We can let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=x}   or  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=\cos(2x).}
In this case, we let    be the polynomial.
So, we let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=x}   and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle dv=\cos(2x)~dx.}
Then,   
To find  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle v,}   we need to use  substitution. So,
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle v=\int \cos(2x)~dx={\frac {1}{2}}\sin(2x).}
Hence, by integration by parts, we get
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {\int x\cos(2x)~dx}&=&\displaystyle {{\frac {1}{2}}x\sin(2x)-\int {\frac {1}{2}}\sin(2x)~dx}\\&&\\&=&\displaystyle {{\frac {1}{2}}x\sin(2x)+{\frac {1}{4}}\cos(2x)+C,}\end{array}}}
where we use   substitution to evaluate the last integral.
Final Answer:  
       Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {1}{2}}x\sin(2x)+{\frac {1}{4}}\cos(2x)+C}

3)   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int \ln x~dx}

Solution:  
We have a choice to make.
We can let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=\ln x}   or  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle dv=\ln x~dx.}
In this case, we don't want to let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle dv=\ln x~dx}   since we don't know how to integrate    yet.
So, we let    and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle dv=1~dx.}
Then,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle du={\frac {1}{x}}~dx}   and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle v=x.}
Hence, by integration by parts, we get
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {\int \ln x~dx}&=&\displaystyle {x\ln(x)-\int x{\bigg (}{\frac {1}{x}}{\bigg )}~dx}\\&&\\&=&\displaystyle {x\ln(x)-\int 1~dx}\\&&\\&=&\displaystyle {x\ln(x)-x+C.}\end{array}}}
Note: The domain of the function    is  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (0,\infty ).}
So, this integral is defined for  
Final Answer:  
       Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x\ln(x)-x+C}

Exercise 1

Evaluate  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int x\sec ^{2}x~dx.}

Since we know the antiderivative of  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sec ^{2}x,}

we let    and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle dv=\sec ^{2}x~dx.}

Then,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle du=dx}   and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle v=\tan x.}

Using integration by parts, we get

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {\int x\sec ^{2}x~dx}&=&\displaystyle {x\tan x-\int \tan x~dx}\\&&\\&=&\displaystyle {x\tan x-\int {\frac {\sin x}{\cos x}}~dx.}\end{array}}}

For the remaining integral, we use  substitution.

Let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=\cos x.}   Then,  

So, we get

So, we have

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int x\sec ^{2}x~dx=x\tan x+\ln |\cos x|+C.}

Exercise 2

Evaluate  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int {\frac {\ln x}{x^{3}}}~dx.}

We start by letting    and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle dv={\frac {1}{x^{3}}}~dx.}

Then,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle du={\frac {1}{x}}~dx}   and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle v=-{\frac {1}{2x^{2}}}.}

So, using integration by parts, we get

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {\int {\frac {\ln x}{x^{3}}}~dx}&=&\displaystyle {-{\frac {\ln x}{2x^{2}}}-\int -{\frac {1}{2x^{2}}}{\bigg (}{\frac {1}{x}}{\bigg )}~dx}\\&&\\&=&\displaystyle {-{\frac {\ln x}{2x^{2}}}+\int {\frac {1}{2x^{3}}}~dx}\\&&\\&=&\displaystyle {-{\frac {\ln x}{2x^{2}}}-{\frac {1}{4x^{2}}}+C.}\end{array}}}

So, we have

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int {\frac {\ln x}{x^{3}}}~dx=-{\frac {\ln x}{2x^{2}}}-{\frac {1}{4x^{2}}}+C.}

Exercise 3

Evaluate  

We start by letting    and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle dv={\sqrt {x+1}}~dx.}

Then,  

To find  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle v,}   we need to use  substitution. So,

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle v=\int {\sqrt {x+1}}~dx={\frac {2}{3}}(x+1)^{\frac {3}{2}}.}

Hence, by integration by parts, we get

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {\int x{\sqrt {x+1}}~dx}&=&\displaystyle {{\frac {2}{3}}x(x+1)^{\frac {3}{2}}-\int {\frac {2}{3}}(x+1)^{\frac {3}{2}}~dx}\\&&\\&=&\displaystyle {{\frac {2}{3}}x(x+1)^{\frac {3}{2}}-{\frac {4}{15}}(x+1)^{\frac {5}{2}}+C,}\end{array}}}

where we use   substitution to evaluate the last integral.

So, we have

Exercise 4

Evaluate  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int x^{2}e^{-2x}~dx.}

We start by letting  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=x^{2}}   and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle dv=e^{-2x}~dx.}

Then,  

To find  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle v,}   we need to use  substitution. So,

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle v=\int e^{-2x}~dx={\frac {e^{-2x}}{-2}}.}

Hence, by integration by parts, we get

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {\int x^{2}e^{-2x}~dx}&=&\displaystyle {{\frac {x^{2}e^{-2x}}{-2}}-\int 2x{\bigg (}{\frac {e^{-2x}}{-2}}{\bigg )}~dx}\\&&\\&=&\displaystyle {{\frac {x^{2}e^{-2x}}{-2}}+\int xe^{-2x}~dx.}\end{array}}}

Now, we need to use integration by parts a second time.

Let    and  

Then,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle du=dx}   and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle v={\frac {e^{-2x}}{-2}}.}

Therefore, using integration by parts again, we get

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {\int x^{2}e^{-2x}~dx}&=&\displaystyle {{\frac {x^{2}e^{-2x}}{-2}}+{\frac {xe^{-2x}}{-2}}-\int {\frac {e^{-2x}}{-2}}~dx}\\&&\\&=&\displaystyle {{\frac {x^{2}e^{-2x}}{-2}}+{\frac {xe^{-2x}}{-2}}+{\frac {e^{-2x}}{-4}}+C.}\end{array}}}

So, we have

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int x^{2}e^{-2x}~dx={\frac {x^{2}e^{-2x}}{-2}}+{\frac {xe^{-2x}}{-2}}+{\frac {e^{-2x}}{-4}}+C.}

Exercise 5

Evaluate  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int e^{3x}\sin(2x)~dx.}

We begin by letting  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=\sin(2x)}   and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle dv=e^{3x}~dx.}

Then,  

To find  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle v,}   we need to use  substitution. So,

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle v=\int e^{3x}~dx={\frac {e^{3x}}{3}}.}

Hence, by integration by parts, we have

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {\int e^{3x}\sin(2x)~dx}&=&\displaystyle {{\frac {e^{3x}\sin(2x)}{3}}-\int {\frac {2}{3}}\cos(2x)e^{3x}~dx}\\&&\\&=&\displaystyle {{\frac {e^{3x}\sin(2x)}{3}}-{\frac {2}{3}}\int \cos(2x)e^{3x}~dx.}\end{array}}}

Now, we need to use integration by parts a second time.

Let  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=\cos (2x)}   and  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle dv=e^{3x}~dx.}

Then,  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du=-2\sin(2x)~dx}   and  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v=\frac{e^{3x}}{3}.}

Therefore, using integration by parts again, we get

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\int e^{3x} \sin(2x)~dx} & = & \displaystyle{\frac{e^{3x}\sin(2x)}{3}-\frac{2}{3} \bigg[ \frac{\cos(2x)e^{3x}}{3}+\int \frac{2}{3}\sin(2x)e^{3x}~dx\bigg]}\\ &&\\ & = & \displaystyle{\frac{e^{3x}\sin(2x)}{3}-\frac{2\cos(2x)e^{3x}}{9}-\frac{4}{9}\int e^{3x}\sin(2x)~dx.} \end{array}}

Now, we have the exact same integral that we had at the beginning of the problem.

So, we add this integral to the other side of the equation.

When we do this, we get

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{13}{9} \int e^{3x}\sin(2x)~dx = \frac{e^{3x}\sin(2x)}{3}-\frac{2\cos(2x)e^{3x}}{9}.}

Therefore, we get

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int e^{3x}\sin(2x)~dx = \frac{9}{13}\bigg(\frac{e^{3x}\sin(2x)}{3}-\frac{2\cos(2x)e^{3x}}{9}\bigg)+C.}

Exercise 6

Evaluate  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \sin(2x)\cos(3x)~dx.}

For this problem, we use a similar process as Exercise 5.

We use integration by parts twice, which produces the same integral given to us in the problem.

Then, we solve for our integral.

We begin by letting  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=\sin(2x)}   and  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle dv=\cos(3x)~dx.}

Then,  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du=2\cos (2x)~dx.}

To find  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v,}   we need to use  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u-} substitution. So,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v=\int \cos(3x)~dx=\frac{1}{3}\sin(3x).}

Hence, by integration by parts, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\int \sin(2x)\cos(3x)~dx} & = & \displaystyle{\frac{1}{3}\sin(2x)\sin(3x)-\int \frac{2}{3}\cos(2x)\sin(3x)~dx}\\ &&\\ & = & \displaystyle{\frac{1}{3}\sin(2x)\sin(3x)-\frac{2}{3} \int \cos(2x)\sin(3x)~dx.} \end{array}}

Now, we need to use integration by parts a second time.

Let  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=\cos (2x)}   and  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle dv=\sin(3x)~dx.}

Then,  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du=-2\sin(2x)~dx}   and  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v=\frac{-\cos(3x)}{3}.}

Therefore, using integration by parts again, we get

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\int \sin(2x)\cos(3x)~dx} & = & \displaystyle{\frac{1}{3}\sin(2x)\sin(3x)-\frac{2}{3} \bigg[ \frac{-\cos(2x)\cos(3x)}{3}-\int \frac{2}{3}\sin(2x)\cos(3x)~dx\bigg]}\\ &&\\ & = & \displaystyle{\frac{1}{3}\sin(2x)\sin(3x)+ \frac{2\cos(2x)\cos(3x)}{9}+\frac{4}{9}\int \sin(2x)\cos(3x)~dx.} \end{array}}

Now, we have the exact same integral that we had at the beginning of the problem.

So, we subtract this integral to the other side of the equation.

When we do this, we get

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{5}{9} \int \sin(2x)\cos(3x)~dx = \frac{1}{3}\sin(2x)\sin(3x)+ \frac{2\cos(2x)\cos(3x)}{9}.}

Therefore, we get

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int e^{3x}\sin(2x)~dx = \frac{9}{5}\bigg(\frac{1}{3}\sin(2x)\sin(3x)+ \frac{2\cos(2x)\cos(3x)}{9}\bigg)+C.}
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