Integration by Parts

Introduction

Let's say we want to integrate

$\int x^{2}e^{x^{3}}~dx.$

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

$\int xe^{x}~dx.$

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

Then,  $du=f'(x)~dx$  and  $v=g(x).$

Plugging these into our formula, we obtain

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

Warm-Up

Evaluate the following integrals.

1)   $\int xe^{x}~dx$

Solution:
We have two options when doing integration by parts.
We can let  $u=x$  or  $u=e^{x}.$
In this case, we let  $u$  be the polynomial.
So, we let  $u=x$  and  $dv=e^{x}~dx.$
Then,  $du=dx$  and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle v=e^{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 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)   $\int x\cos(2x)~dx$

Solution:
We have a choice to make.
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=\cos(2x).}
In this case, we let  $u$  be the polynomial.
So, we let  $u=x$  and  $dv=\cos(2x)~dx.$
Then,  $du=dx.$
By  $u-$substitution,
$v=\int \cos(2x)~dx={\frac {1}{2}}\sin(2x).$
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  $u=\ln x$  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  $\ln x$  yet.
So, we let  $u=\ln x$  and  $dv=1~dx.$
Then,  $du={\frac {1}{x}}~dx$  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
${\begin{array}{rcl}\displaystyle {\int \ln x~dx}&=&\displaystyle {x\ln(x)-\int x{\bigg (}{\frac {1}{x}}{\bigg )}~dx}\\&&\\&=&\displaystyle {x\ln(x)-\int 1~dx}\\&&\\&=&\displaystyle {x\ln(x)-x+C.}\end{array}}$

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  $u=\cos x.$  Then,  $du=-\sin x~dx.$

So, we get

${\begin{array}{rcl}\displaystyle {\int x\sec ^{2}x~dx}&=&\displaystyle {x\tan x+\int {\frac {1}{u}}~du}\\&&\\&=&\displaystyle {x\tan x+\ln |u|+C}\\&&\\&=&\displaystyle {x\tan x+\ln |\cos x|+C.}\end{array}}$

So, we have

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  $\int {\frac {\ln x}{x^{3}}}~dx.$

We start by letting  $u=\ln x$  and  $dv={\frac {1}{x^{3}}}~dx.$

Then,  $du={\frac {1}{x}}~dx$  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

${\begin{array}{rcl}\displaystyle {\int {\frac {\ln x}{x^{3}}}~dx}&=&\displaystyle {-{\frac {\ln x}{2x^{2}}}-\int -{\frac {1}{2x^{2}}}{\bigg (}{\frac {1}{x}}{\bigg )}~dx}\\&&\\&=&\displaystyle {-{\frac {\ln x}{2x^{2}}}+\int {\frac {1}{2x^{3}}}~dx}\\&&\\&=&\displaystyle {-{\frac {\ln x}{2x^{2}}}-{\frac {1}{4x^{2}}}+C.}\end{array}}$

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  $\int x{\sqrt {x+1}}~dx.$

We start by letting  $u=x$  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,  $du=dx.$

Also, by  $u-$substitution, we have

$v=\int {\sqrt {x+1}}~dx={\frac {2}{3}}(x+1)^{\frac {3}{2}}.$

Hence, by integration by parts, we get

${\begin{array}{rcl}\displaystyle {\int x{\sqrt {x+1}}~dx}&=&\displaystyle {{\frac {2}{3}}x(x+1)^{\frac {3}{2}}-\int {\frac {2}{3}}(x+1)^{\frac {3}{2}}~dx}\\&&\\&=&\displaystyle {{\frac {2}{3}}x(x+1)^{\frac {3}{2}}-{\frac {4}{15}}(x+1)^{\frac {5}{2}}+C,}\end{array}}$

where we use  $u-$ substitution to evaluate the last integral.

So, we have

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

Exercise 4

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

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

Also, by  $u-$substitution, we have

$v=\int e^{-2x}~dx={\frac {e^{-2x}}{-2}}.$

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

Then,  $du=dx$  and  $v={\frac {e^{-2x}}{-2}}.$

Therefore, using integration by parts again, we get

${\begin{array}{rcl}\displaystyle {\int x^{2}e^{-2x}~dx}&=&\displaystyle {{\frac {x^{2}e^{-2x}}{-2}}+{\frac {xe^{-2x}}{-2}}-\int {\frac {e^{-2x}}{-2}}~dx}\\&&\\&=&\displaystyle {{\frac {x^{2}e^{-2x}}{-2}}+{\frac {xe^{-2x}}{-2}}+{\frac {e^{-2x}}{-4}}+C.}\end{array}}$

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  $\int e^{3x}\sin(2x)~dx.$

We begin by letting  $u=\sin(2x)$  and  $dv=e^{3x}~dx.$

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

Also, by  $u-$ substitution, we have

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

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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=\cos(2x)}   and  $dv=e^{3x}~dx.$

Then,  $du=-2\sin(2x)~dx$  and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle v={\frac {e^{3x}}{3}}.}

Therefore, using integration by parts again, we get

${\begin{array}{rcl}\displaystyle {\int e^{3x}\sin(2x)~dx}&=&\displaystyle {{\frac {e^{3x}\sin(2x)}{3}}-{\frac {2}{3}}{\bigg [}{\frac {\cos(2x)e^{3x}}{3}}+\int {\frac {2}{3}}\sin(2x)e^{3x}~dx{\bigg ]}}\\&&\\&=&\displaystyle {{\frac {e^{3x}\sin(2x)}{3}}-{\frac {2\cos(2x)e^{3x}}{9}}-{\frac {4}{9}}\int e^{3x}\sin(2x)~dx.}\end{array}}$

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

${\frac {13}{9}}\int e^{3x}\sin(2x)~dx={\frac {e^{3x}\sin(2x)}{3}}-{\frac {2\cos(2x)e^{3x}}{9}}.$

Therefore, we get

$\int e^{3x}\sin(2x)~dx={\frac {9}{13}}{\bigg (}{\frac {e^{3x}\sin(2x)}{3}}-{\frac {2\cos(2x)e^{3x}}{9}}{\bigg )}+C.$

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.}

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 u=2\cos (2x)~dx.}

Also, 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, 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 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}+\int \frac{4}{9}\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.}