Integration by Parts

Introduction

Let's say we want to integrate

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^{x^3}~dx.}

Here, we can compute this antiderivative by using  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.

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

For example, consider the integral

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 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,

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

Then, 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{f(x)g(x)} & = & \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 (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 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 (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=f(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=g'(x)~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=f'(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=g(x).}

Plugging these into our formula, we obtain

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 u~dv=uv-\int v~du.}

Warm-Up

Evaluate the following integrals.

1)   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 xe^x~dx}

Solution:
We have two options when doing integration by parts.
We can 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}   or  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=e^x.}
In this case, we 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}   be the polynomial.
So, we 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^x~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=e^x.}
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 xe^x~dx} & = & \displaystyle{xe^x-\int e^x~dx}\\ &&\\ & = & \displaystyle{xe^x-e^x+C.} \end{array}}

Final Answer:
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 xe^x-e^x+C}

2)   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\cos (2x)~dx}

Solution:
We have a choice to make.
We can 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}   or  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).}
In this case, we 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}   be the polynomial.
So, we 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=\cos(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.}
By  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,  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(2x)~dx=\frac{1}{2} \sin(2x).}
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\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  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.

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

3)   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 \ln x~dx}

Solution:
We have a choice to make.
We can 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=\ln x}   or  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=\ln x~dx.}
In this case, we don't want to 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 dv=\ln x~dx}   since we don't know how to integrate  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 \ln x}   yet.
So, we 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=\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=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=\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=x.}
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 \ln x~dx} & = & \displaystyle{x\ln (x) -\int x \big(\frac{1}{x}\big)~dx}\\ &&\\ & = & \displaystyle{x\ln (x)-\int 1~dx}\\ &&\\ & = & \displaystyle{x\ln (x)-x+C.} \end{array}}

Final Answer:
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 x\ln (x)-x+C}

Exercise 1

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^3 e^{-2x}~dx.}

First, we need to know the derivative of  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 \csc x.}   Recall

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 \csc x =\frac{1}{\sin x}.}

Now, using the Quotient Rule, 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{\frac{d}{dx}(\csc x)} & = & \displaystyle{\frac{d}{dx}\bigg(\frac{1}{\sin x}\bigg)}\\ &&\\ & = & \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.} \end{array}}

Using the Product Rule and Power Rule, 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{f'(x)} & = & \displaystyle{\frac{1}{x^2}(\csc x-4)'+\bigg(\frac{1}{x^2}\bigg)'(\csc x-4)}\\ &&\\ & = & \displaystyle{\frac{1}{x^2}(-\csc x \cot x+0)+(-2x^{-3})(\csc x-4)}\\ &&\\ & = & \displaystyle{\frac{-\csc x \cot x}{x^2}+\frac{-2(\csc x-4)}{x^3}.} \end{array}}

So, we have

$f'(x)={\frac {-\csc x\cot x}{x^{2}}}+{\frac {-2(\csc x-4)}{x^{3}}}.$

Exercise 2

Evaluate  $\int e^{3x}\sin(2x)~dx.$

Notice that the function  $g(x)$  is the product of three functions.

We start by grouping two of the functions together. So, we have  $g(x)=(2x\sin x)\sec x.$

Using the Product Rule, we get

${\begin{array}{rcl}\displaystyle {g'(x)}&=&\displaystyle {(2x\sin x)(\sec x)'+(2x\sin x)'\sec x}\\&&\\&=&\displaystyle {(2x\sin x)(\tan ^{2}x)+(2x\sin x)'\sec x.}\end{array}}$

Now, we need to use the Product Rule again. So,

${\begin{array}{rcl}\displaystyle {g'(x)}&=&\displaystyle {2x\sin x\tan ^{2}x+(2x(\sin x)'+(2x)'\sin x)\sec x}\\&&\\&=&\displaystyle {2x\sin x\tan ^{2}x+(2x\cos x+2\sin x)\sec x.}\end{array}}$

So, we have

$g'(x)=2x\sin x\tan ^{2}x+(2x\cos x+2\sin x)\sec x.$

But, there is another way to do this problem. Notice

${\begin{array}{rcl}\displaystyle {g(x)}&=&\displaystyle {2x\sin x\sec x}\\&&\\&=&\displaystyle {2x\sin x{\frac {1}{\cos x}}}\\&&\\&=&\displaystyle {2x\tan x.}\end{array}}$

Now, you would only need to use the Product Rule once instead of twice.

Exercise 3

Evaluate  $\int x\sec ^{2}x~dx.$

Using the Quotient Rule, we have

$h'(x)={\frac {(x^{2}\cos x+3)(x^{2}\sin x+1)'-(x^{2}\sin x+1)(x^{2}\cos x+3)'}{(x^{2}\cos x+3)^{2}}}.$

Now, we need to use the Product Rule. So, we have

${\begin{array}{rcl}\displaystyle {h'(x)}&=&\displaystyle {\frac {(x^{2}\cos x+3)(x^{2}(\sin x)'+(x^{2})'\sin x)-(x^{2}\sin x+1)(x^{2}(\cos x)'+(x^{2})'\cos x)}{(x^{2}\cos x+3)^{2}}}\\&&\\&=&\displaystyle {{\frac {(x^{2}\cos x+3)(x^{2}\cos x+2x\sin x)-(x^{2}\sin x+1)(-x^{2}\sin x+2x\cos x)}{(x^{2}\cos x+3)^{2}}}.}\end{array}}$

So, we get

$h'(x)={\frac {(x^{2}\cos x+3)(x^{2}\cos x+2x\sin x)-(x^{2}\sin x+1)(-x^{2}\sin x+2x\cos x)}{(x^{2}\cos x+3)^{2}}}.$

Exercise 4

Evaluate  $\int x{\sqrt {x+1}}~dx.$

First, using the Quotient Rule, we have

${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\frac {x^{2}\sin x(e^{x})'-e^{x}(x^{2}\sin x)'}{(x^{2}\sin x)^{2}}}\\&&\\&=&\displaystyle {{\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}\sin x)'}{x^{4}\sin ^{2}x}}.}\end{array}}$

Now, we need to use the Product Rule. So, we have

${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}(\sin x)'+(x^{2})'\sin x)}{x^{4}\sin ^{2}x}}\\&&\\&=&\displaystyle {{\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}\cos x+2x\sin x)}{x^{4}\sin ^{2}x}}.}\end{array}}$

So, we have

$f'(x)={\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}\cos x+2x\sin x)}{x^{4}\sin ^{2}x}}.$

Exercise 5

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

First, using the Quotient Rule, we have

${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\frac {x^{2}\sin x(e^{x})'-e^{x}(x^{2}\sin x)'}{(x^{2}\sin x)^{2}}}\\&&\\&=&\displaystyle {{\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}\sin x)'}{x^{4}\sin ^{2}x}}.}\end{array}}$

Now, we need to use the Product Rule. So, we have

${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}(\sin x)'+(x^{2})'\sin x)}{x^{4}\sin ^{2}x}}\\&&\\&=&\displaystyle {{\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}\cos x+2x\sin x)}{x^{4}\sin ^{2}x}}.}\end{array}}$

So, we have

$f'(x)={\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}\cos x+2x\sin x)}{x^{4}\sin ^{2}x}}.$

Exercise 6

Evaluate  $\int \sin(2x)\cos(3x)~dx.$

First, using the Quotient Rule, we have

${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\frac {x^{2}\sin x(e^{x})'-e^{x}(x^{2}\sin x)'}{(x^{2}\sin x)^{2}}}\\&&\\&=&\displaystyle {{\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}\sin x)'}{x^{4}\sin ^{2}x}}.}\end{array}}$

Now, we need to use the Product Rule. So, we have

${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}(\sin x)'+(x^{2})'\sin x)}{x^{4}\sin ^{2}x}}\\&&\\&=&\displaystyle {{\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}\cos x+2x\sin x)}{x^{4}\sin ^{2}x}}.}\end{array}}$

So, we have

$f'(x)={\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}\cos x+2x\sin x)}{x^{4}\sin ^{2}x}}.$