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 (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 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 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))'~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 (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.}
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(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 \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 (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} is 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 (0,\infty).}
So, this integral is defined for 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>0.}

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 \sec^2 x~dx.}

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

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=\sec^2 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=\tan x.}

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 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 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.

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 x.} 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=-\sin x~dx.}

So, 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 \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 (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 \sec^2 x~dx=x\tan x+ \ln |\cos x|+C.}

Exercise 2

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 \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 $dv=e^{-2x}~dx.$

Then, $du=2x~dx.$

To find $v,$ we need to use $u-$ substitution. So,

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

Hence, by integration by parts, we get

{\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 $dv=e^{-2x}~dx.$

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

\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, $du=2\cos(2x)~dx.$

To find $v,$ we need to use $u-$ substitution. So,

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

Hence, by integration by parts, we have

{\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 $u=\cos(2x)$ and $dv=e^{3x}~dx.$

Then, $du=-2\sin(2x)~dx$ and $v={\frac {e^{3x}}{3}}.$

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 $\int \sin(2x)\cos(3x)~dx.$

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 $u=\sin(2x)$ and $dv=\cos(3x)~dx.$

Then, $du=2\cos(2x)~dx.$

To find $v,$ we need to use $u-$ substitution. So,

v=\int \cos(3x)~dx={\frac {1}{3}}\sin(3x).

Hence, by integration by parts, we have

{\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 $u=\cos(2x)$ and $dv=\sin(3x)~dx.$

Then, $du=-2\sin(2x)~dx$ and $v={\frac {-\cos(3x)}{3}}.$

Therefore, using integration by parts again, we get

{\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

{\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

\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

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@@ 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 193: / Line 202: @@
 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>
-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>
+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>
 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}\big(\frac{1}{x}\big)~dx}\\
+\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}+\int \frac{1}{2x^3}~dx}\\
 &&\\
-& = & \displaystyle{\frac{-\ln x}{2x^2}-\frac{1}{4x^2}+C.}
+& = & \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>
+::<math>\int \frac{\ln x}{x^3}~dx=-\frac{\ln x}{2x^2}-\frac{1}{4x^2}+C.</math>
 == Exercise 3 ==
+Evaluate &nbsp;<math style="vertical-align: -13px">\int x\sqrt{x+1}~dx.</math>
+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>
-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}
-\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 278: / 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>