Difference between revisions of "Integration by Parts"

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 $u-$ substitution.

While $u-$ substitution is an important integration technique, it will not help us evaluate all integrals.

For example, consider the integral

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

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,

(f(x)g(x))'=f'(x)g(x)+f(x)g'(x).

Then, we have

{\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

\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 $u=f(x)$ and $dv=g'(x)~dx.$

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

\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 $u=x$ 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 $u$ be the polynomial.
So, we let $u=x$ and $dv=e^{x}~dx.$
Then, $du=dx$ and $v=e^{x}.$
Hence, by integration by parts, we get
${\begin{array}{rcl}\displaystyle {\int xe^{x}~dx}&=&\displaystyle {xe^{x}-\int e^{x}~dx}\\&&\\&=&\displaystyle {xe^{x}-e^{x}+C.}\end{array}}$

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 $u=x$ or $u=\cos(2x).$
In this case, we let $u$ be the polynomial.
So, we let $u=x$ and $dv=\cos(2x)~dx.$
Then, $du=dx.$
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 $u-$ substitution. So,
$v=\int \cos(2x)~dx={\frac {1}{2}}\sin(2x).$
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 $u-$ substitution to evaluate the last integral.

Final Answer:
${\frac {1}{2}}x\sin(2x)+{\frac {1}{4}}\cos(2x)+C$

3) $\int \ln x~dx$

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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \ln x} yet.
So, we let Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=\ln x} and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle dv=1~dx.}
Then, $du={\frac {1}{x}}~dx$ and $v=x.$
Hence, by integration by parts, we get
${\begin{array}{rcl}\displaystyle {\int \ln x~dx}&=&\displaystyle {x\ln(x)-\int x{\bigg (}{\frac {1}{x}}{\bigg )}~dx}\\&&\\&=&\displaystyle {x\ln(x)-\int 1~dx}\\&&\\&=&\displaystyle {x\ln(x)-x+C.}\end{array}}$
Note: The domain of the function Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \ln x} 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 $x>0.$

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 $\int x\sec ^{2}x~dx.$

Since we know the antiderivative of $\sec ^{2}x,$

we let $u=x$ and $dv=\sec ^{2}x~dx.$

Then, $du=dx$ 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 $u-$ substitution.

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

So, 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 {\frac {1}{u}}~du}\\&&\\&=&\displaystyle {x\tan x+\ln |u|+C}\\&&\\&=&\displaystyle {x\tan x+\ln |\cos x|+C.}\end{array}}}

So, we have

\int x\sec ^{2}x~dx=x\tan x+\ln |\cos x|+C.

Exercise 2

Evaluate $\int {\frac {\ln x}{x^{3}}}~dx.$

We start 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=\ln x} 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=\frac{1}{x^3}~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=\frac{1}{x}~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{1}{2x^2}.}

So, using integration by parts, 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 \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 (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 \frac{\ln x}{x^3}~dx=-\frac{\ln x}{2x^2}-\frac{1}{4x^2}+C.}

Exercise 3

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 x\sqrt{x+1}~dx.}

We start 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=x} 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=\sqrt{x+1}~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=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 \sqrt{x+1}~dx=\frac{2}{3}(x+1)^{\frac{3}{2}}.}

Hence, by integration by parts, 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 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 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 to evaluate the last integral.

So, 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 \int x\sqrt{x+1}~dx=\frac{2}{3}x(x+1)^{\frac{3}{2}}-\frac{4}{15}(x+1)^{\frac{5}{2}}+C.}

Exercise 4

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 x^2e^{-2x}~dx.}

We start 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=x^2} 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^{-2x}~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=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 e^{-2x}~dx=\frac{e^{-2x}}{-2}.}

Hence, by integration by parts, 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 x^2 e^{-2x}~dx} & = & \displaystyle{\frac{x^2e^{-2x}}{-2}-\int 2x\bigg(\frac{e^{-2x}}{-2}\bigg)~dx}\\ &&\\ & = & \displaystyle{\frac{x^2e^{-2x}}{-2}+\int xe^{-2x}~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=x} 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^{-2x}~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=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^{-2x}}{-2}.}

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 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^2e^{-2x}}{-2}+\frac{xe^{-2x}}{-2}+\frac{e^{-2x}}{-4}+C.} \end{array}}

So, 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 \int x^2e^{-2x}~dx=\frac{x^2e^{-2x}}{-2}+\frac{xe^{-2x}}{-2}+\frac{e^{-2x}}{-4}+C.}

Exercise 5

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 e^{3x}\sin(2x)~dx.}

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=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\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 e^{3x}~dx=\frac{e^{3x}}{3}.}

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 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.}

Difference between revisions of "Integration by Parts"

Latest revision as of 12:17, 30 October 2017

Contents

Introduction

Warm-Up

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Exercise 6

Navigation menu

Search

@@ Line 25: / Line 25: @@
 ::<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.}
@@ Line 98: / Line 100: @@
 |Then, &nbsp;<math style="vertical-align: 0px">du=dx.</math>&nbsp;
 |-
-|By &nbsp;<math style="vertical-align: 0px">u-</math>substitution, &nbsp;<math style="vertical-align: -13px">v=\int \cos(2x)~dx=\frac{1}{2} \sin(2x).</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: -13px">v=\int \cos(2x)~dx=\frac{1}{2} \sin(2x).</math>
 |-
 |Hence, by integration by parts, we get
@@ Line 137: / Line 142: @@
 |
 ::<math>\begin{array}{rcl}
-\displaystyle{\int \ln x~dx} & = & \displaystyle{x\ln (x) -\int x \big(\frac{1}{x}\big)~dx}\\
+\displaystyle{\int \ln x~dx} & = & \displaystyle{x\ln (x) -\int x \bigg(\frac{1}{x}\bigg)~dx}\\
 &&\\
 & = & \displaystyle{x\ln (x)-\int 1~dx}\\
@@ Line 143: / Line 148: @@
 & = & \displaystyle{x\ln (x)-x+C.}
 \end{array}</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>
+|-
+|So, this integral is defined for &nbsp;<math style="vertical-align: 0px">x>0.</math>
 |}
@@ Line 189: / Line 198: @@
 == 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}
-\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>
 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.
-== Exercise 3 ==
+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>
-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}
-\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>
-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 ==
-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>
+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>
-First, using the Quotient Rule, we have
+Hence, by integration by parts, we get
 ::<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>
-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}
-\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>
 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 ==
-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}
-\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>
-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}
-\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>
-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 ==
@@ Line 295: / Line 330: @@
 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}
-\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>
-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}
-\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>
-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>