MathAdmin: Created page with "==Introduction== Let's say we want to integrate :: $\int x^2e^{x^3}~dx.$ Here, we can compute this antiderivative by using $..."$

2017-10-30T19:20:44Z

Created page with "==Introduction== Let's say we want to integrate ::<math>\int x^2e^{x^3}~dx.</math> Here, we can compute this antiderivative by using <math style="vertical-align: 0px">..."

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==Introduction==
Let's say we want to integrate

::<math>\int x^2e^{x^3}~dx.</math>

Here, we can compute this antiderivative by using  <math style="vertical-align: 0px">u-</math>substitution.

While  <math style="vertical-align: 0px">u-</math>substitution is an important integration technique, it will not help us evaluate all integrals.

For example, consider the integral

::<math>\int xe^x~dx.</math>

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,

::<math>(f(x)g(x))'=f'(x)g(x)+f(x)g'(x).</math>

Then, we have

::<math>\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}</math>

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

::<math>\int f(x)g'(x)~dx = f(x)g(x)-\int f'(x)g(x)~dx.</math>

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  <math style="vertical-align: -5px">u=f(x)</math>  and  <math style="vertical-align: -5px">dv=g'(x)~dx.</math>

Then,  <math style="vertical-align: -5px">du=f'(x)~dx</math>  and  <math style="vertical-align: -5px">v=g(x).</math>

Plugging these into our formula, we obtain

::<math>\int u~dv=uv-\int v~du.</math>

==Warm-Up==
Evaluate the following integrals.

'''1)'''   <math style="vertical-align: -13px">\int xe^x~dx</math>

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution:  
|-
|We have two options when doing integration by parts.
|-
|We can let  <math style="vertical-align: 0px">u=x</math>  or  <math style="vertical-align: 0px">u=e^x.</math>
|-
|In this case, we let  <math style="vertical-align: 0px">u</math>  be the polynomial.
|-
|So, we let  <math style="vertical-align: 0px">u=x</math>  and  <math style="vertical-align: 0px">dv=e^x~dx.</math>
|-
|Then,  <math style="vertical-align: 0px">du=dx</math>  and  <math style="vertical-align: 0px">v=e^x.</math>
|-
|Hence, by integration by parts, we get
|-
|
::<math>\begin{array}{rcl}
\displaystyle{\int xe^x~dx} & = & \displaystyle{xe^x-\int e^x~dx}\\
&&\\
& = & \displaystyle{xe^x-e^x+C.}
\end{array}</math>
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|       <math>xe^x-e^x+C</math>
|}

'''2)'''   <math style="vertical-align: -14px">\int x\cos (2x)~dx</math>

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution:  
|-
|
We have a choice to make.
|-
|We can let  <math style="vertical-align: 0px">u=x</math>  or  <math style="vertical-align: -5px">u=\cos(2x).</math>
|-
|In this case, we let  <math style="vertical-align: 0px">u</math>  be the polynomial.
|-
|So, we let  <math style="vertical-align: 0px">u=x</math>  and  <math style="vertical-align: -5px">dv=\cos(2x)~dx.</math>
|-
|Then,  <math style="vertical-align: 0px">du=dx.</math> 
|-
|To find  <math style="vertical-align: -4px">v,</math>  we need to use  <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
|-
|
::<math>\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}</math>
|-
|where we use  <math style="vertical-align: 0px">u-</math> substitution to evaluate the last integral.
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|       <math>\frac{1}{2}x\sin(2x)+\frac{1}{4}\cos(2x)+C</math>
|}

'''3)'''   <math style="vertical-align: -14px">\int \ln x~dx</math>

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution:  
|-
|We have a choice to make.
|-
|We can let  <math style="vertical-align: -1px">u=\ln x</math>  or  <math style="vertical-align: -1px">dv=\ln x~dx.</math>
|-
|In this case, we don't want to let  <math style="vertical-align: -1px">dv=\ln x~dx</math>  since we don't know how to integrate  <math style="vertical-align: -1px">\ln x</math>  yet.
|-
|So, we let  <math style="vertical-align: -1px">u=\ln x</math>  and  <math style="vertical-align: -1px">dv=1~dx.</math>
|-
|Then,  <math style="vertical-align: -13px">du=\frac{1}{x}~dx</math>  and  <math style="vertical-align: 0px">v=x.</math>
|-
|Hence, by integration by parts, we get
|-
|
::<math>\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}</math>
|-
|<u>Note:</u> The domain of the function  <math style="vertical-align: -1px">\ln x</math>  is  <math style="vertical-align: -5px">(0,\infty).</math>
|-
|So, this integral is defined for  <math style="vertical-align: 0px">x>0.</math>
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|       <math>x\ln (x)-x+C</math>
|-
|}

== Exercise 1 ==

Evaluate  <math style="vertical-align: -13px">\int x \sec^2 x~dx.</math>

Since we know the antiderivative of  <math style="vertical-align: -3px">\sec^2 x,</math>

we let  <math style="vertical-align: 0px">u=x</math>  and  <math style="vertical-align: 0px">dv=\sec^2 x~dx.</math>

Then,  <math style="vertical-align: 0px">du=dx</math>  and  <math style="vertical-align: -1px">v=\tan x.</math>

Using integration by parts, we get

::<math>\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}</math>

For the remaining integral, we use  <math style="vertical-align: 0px">u-</math>substitution.

Let  <math style="vertical-align: 0px">u=\cos x.</math>  Then,  <math style="vertical-align: -1px">du=-\sin x~dx.</math>

So, we get

::<math>\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}</math>

So, we have
::<math>\int x \sec^2 x~dx=x\tan x+ \ln |\cos x|+C.</math>

== Exercise 2 ==

Evaluate  <math style="vertical-align: -13px">\int \frac{\ln x}{x^3}~dx.</math>

We start by letting  <math style="vertical-align: -1px">u=\ln x</math>  and  <math style="vertical-align: -13px">dv=\frac{1}{x^3}~dx.</math>

Then,  <math style="vertical-align: -13px">du=\frac{1}{x}~dx</math>  and  <math style="vertical-align: -13px">v=-\frac{1}{2x^2}.</math>

So, using integration by parts, we get

::<math>\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}</math>

So, we have

::<math>\int \frac{\ln x}{x^3}~dx=-\frac{\ln x}{2x^2}-\frac{1}{4x^2}+C.</math>

== Exercise 3 ==
Evaluate  <math style="vertical-align: -13px">\int x\sqrt{x+1}~dx.</math>

We start by letting  <math style="vertical-align: 0px">u=x</math>  and  <math style="vertical-align: -3px">dv=\sqrt{x+1}~dx.</math>

Then,  <math style="vertical-align: 0px">du=dx.</math>

To find  <math style="vertical-align: -4px">v,</math>  we need to use  <math style="vertical-align: 0px">u-</math>substitution. So,

::<math style="vertical-align: 0px">v=\int \sqrt{x+1}~dx=\frac{2}{3}(x+1)^{\frac{3}{2}}.</math>

Hence, by integration by parts, we get

::<math>\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}</math>

where we use  <math style="vertical-align: 0px">u-</math> substitution to evaluate the last integral.

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  <math style="vertical-align: -13px">\int x^2e^{-2x}~dx.</math>

We start by letting  <math style="vertical-align: 0px">u=x^2</math>  and  <math style="vertical-align: 0px">dv=e^{-2x}~dx.</math>

Then,  <math style="vertical-align: 0px">du=2x~dx.</math>

To find  <math style="vertical-align: -4px">v,</math>  we need to use  <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}
\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}</math>

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

Let  <math style="vertical-align: 0px">u=x</math>  and  <math style="vertical-align: 0px">dv=e^{-2x}~dx.</math>

Then,  <math style="vertical-align: 0px">du=dx</math>  and  <math style="vertical-align: -13px">v=\frac{e^{-2x}}{-2}.</math>

Therefore, using integration by parts again, we get

::<math>\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}</math>

So, we have
::<math>\int x^2e^{-2x}~dx=\frac{x^2e^{-2x}}{-2}+\frac{xe^{-2x}}{-2}+\frac{e^{-2x}}{-4}+C.</math>

== Exercise 5 ==

Evaluate  <math style="vertical-align: -13px">\int e^{3x}\sin(2x)~dx.</math>

We begin by letting  <math style="vertical-align: -6px">u=\sin(2x)</math>  and  <math style="vertical-align: 0px">dv=e^{3x}~dx.</math>

Then,  <math style="vertical-align: -5px">du=2\cos (2x)~dx.</math>

To find  <math style="vertical-align: -4px">v,</math>  we need to use  <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{\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}</math>

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

Let  <math style="vertical-align: -5px">u=\cos (2x)</math>  and  <math style="vertical-align: 0px">dv=e^{3x}~dx.</math>

Then,  <math style="vertical-align: -5px">du=-2\sin(2x)~dx</math>  and  <math style="vertical-align: -13px">v=\frac{e^{3x}}{3}.</math>

Therefore, using integration by parts again, we get

::<math>\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}</math>

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

::<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 ==

Evaluate  <math style="vertical-align: -14px">\int \sin(2x)\cos(3x)~dx.</math>

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  <math style="vertical-align: -5px">u=\sin(2x)</math>  and  <math style="vertical-align: -5px">dv=\cos(3x)~dx.</math>

Then,  <math style="vertical-align: -5px">du=2\cos (2x)~dx.</math>

To find  <math style="vertical-align: -4px">v,</math>  we need to use  <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{\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}</math>

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

Let  <math style="vertical-align: -5px">u=\cos (2x)</math>  and  <math style="vertical-align: -5px">dv=\sin(3x)~dx.</math>

Then,  <math style="vertical-align: -5px">du=-2\sin(2x)~dx</math>  and  <math style="vertical-align: -13px">v=\frac{-\cos(3x)}{3}.</math>

Therefore, using integration by parts again, we get

::<math>\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}</math>

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

::<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>

Integration by Parts - Revision history

MathAdmin: Created page with "==Introduction== Let's say we want to integrate ::\int x^2e^{x^3}~dx. Here, we can compute this antiderivative by using ..."

MathAdmin: Created page with "==Introduction== Let's say we want to integrate :: $\int x^2e^{x^3}~dx.$ Here, we can compute this antiderivative by using $..."$