Integration by Parts: Problems with Solutions

Solution:
We have to solve $\int x^{3}\ln $ $x$ $dx$
$u=\ln $ $x,$ $dv=x^{3}dx.$

Then $du=\frac{dx}{x}\qquad v=\int x^{3}dx=\frac{x^{4}}{4}$ and now we apply

$\int udv=uv-\int vdu=\frac{x^{4}}{4}\ln $ $x-\int \frac{x^{4}}{4}\frac{dx}{x}=\frac{x^{4}}{4}\ln $ $x-\frac{1}{4}\int x^{3}dx$

So $\int x^{3}\ln $ $x$ $dx=\frac{x^{4}}{4}\ln $ $x-\frac{1}{16}x^{4}+C=$
$\frac{x^{4}}{4}\left(\ln x-\frac{1}{4}\right) +C$

Solution:
If $u=4x+7,$ $dv=e^{x}dx$ then $du=4\qquad v=\int e^{x}dx=e^{x}$

so $\int udv=uv-\int vdu$ then $\int (4x+7)e^{x}dx=\left( 4x+7\right) e^{x}-4\int e^{x}dx$

$\int (4x+7)e^{x}dx=\left( 4x+7\right) e^{x}-4e^{x}+C=\left( 4x+3\right) e^{x}+C$

Solution:
Let $u=x\qquad dv=\sin 3x$ $dx$ then
$du=dx\qquad v=\int \sin 3x$ $dx$
$v=-\frac{1}{3}\cos 3x$ so $\int udv=uv-\int vdu$ and replacing

$\int x\sin 3x$ $dx=-\frac{1}{3}x\cos 3x-\int -\frac{1}{3}\cos 3xdx=$

$=-\frac{1}{3}x\cos 3x+\frac{1}{3}\int \cos 3xdx=-\frac{1}{3}x\cos 3x+\frac{1}{9}\sin 3x+C$

Solution:
When we do $u=x,$ $dv=\cos 4x\ dx$ we obtain that
$du=dx\qquad v=\int \cos 4x$ $dx=\frac{1}{4}\sin 4x$
then if we use the formula $\int udv=uv-\int vdu$


$\int x\cos 4x\ dx=\frac{1}{4}x\sin 4x-\frac{1}{4}\int \sin 4xdx$ then

$\int x\cos 4x\ dx=\frac{1}{4}x\sin 4x+\frac{1}{16}\cos 4x+C$

Solution:
$\int \cos ^{3}x\sin x$ $dx=\int \cos ^{2}x\sin x\cos x$ $dx$ and now we do
$\cos ^{2}x=1-\sin ^{2}x$ so $\int \cos ^{3}x\sin x$ $dx=$
$=\int \left( 1-\sin ^{2}x\right) \sin x\cos x$ $dx=\int \left( \sin x-\sin ^{3}x\right) \cos x$ $dx$

and now $u=\sin x\Rightarrow du=\cos xdx$ then

$\int \cos ^{3}x\sin x$ $dx=\int udu-\int u^{3}du=\frac{u^{2}}{2}-\frac{u^{4}}{4}+C$

then $\int \cos ^{3}x\sin x$ $dx=\frac{\sin ^{2}x}{2}-\frac{\sin ^{4}x}{4}+C$

Solution:
$\int \cos^{3}x\sin^{4}x\ dx=\int \cos^{2}x\sin^{4}x\cos x$ $dx$ and now we do

$\cos ^{2}x=1-\sin ^{2}x$
So $\int \cos^{3}x\sin^{4}x$ $dx=$
$=\int \left( 1-\sin ^{2}x\right) \sin ^{4}x\cos x$ $dx=\int \left( \sin^{4}x-\sin ^{6}x\right) \cos x\ dx$

and now $u=\sin x\Rightarrow du=\cos xdx$ then
$\int \cos ^{3}x\sin ^{4}x$ $dx=\int u^{4}du-\int u^{6}du=\frac{u^{5}}{5}-\frac{u^{7}}{7}+C$
then $\int \cos ^{3}x\sin ^{4}x$ $dx=\frac{\sin ^{5}x}{5}-\frac{\sin ^{7}x}{7}+C$

Solution:
$\int \sin ^{3}x\cos ^{2}x\ dx=\int \sin ^{2}x\sin x\cos ^{2}x\ dx$ and since
$\sin ^{2}x=1-\cos ^{2}x$ we get $\int \sin ^{3}x\cos ^{2}x\ dx=$
$=\int \left( 1-\cos ^{2}x\right) \sin x\cos ^{2}xdx=\int \left( \cos^{2}x-\cos ^{4}x\right) \sin xdx$

Now we can do $u=\cos x\qquad du=-\sin xdx$ so

$\int \sin ^{3}x\cos ^{2}x$ $dx=-\int u^{2}du+\int u^{4}du=-\frac{u^{3}}{3}+ \frac{u^{5}}{5}+C$

then $\int \sin ^{3}x\cos ^{2}x$ $dx=-\frac{\cos ^{3}x}{3}+\frac{\cos ^{5}x}{5}+C$

Solution:
We can do, $\int \sin ^{3}x\ dx=\int \sin ^{2}x\sin x$ $dx=\int \left( 1-\cos ^{2}x\right) \sin x\ dx$

then $\int \sin ^{3}x$ $dx=\int \sin xdx-\int \cos ^{2}x\sin x\ dx$ and
substitute $u=\cos x\qquad du=-\sin xdx$ so

$\int \sin ^{3}x$ $dx=-\cos x+\int u^{2}du=-\cos x+\frac{u^{3}}{3}+C$

Then $\int \sin ^{3}x$ $dx=$ $-\cos x+\frac{\cos ^{3}x}{3}+C$

Solution:
Substitute $u=x\qquad dv=e^{-2x}\ dx$
So $du=dx\qquad v=\int e^{-2x}$ $dx=-\frac{1}{2}e^{-2x}$ then we apply

$\int udv=uv-\int vdu\Rightarrow$
$\int xe^{-2x}$ $dx=-\frac{1}{2}xe^{-2x}-\int -\frac{1}{2}e^{-2x}dx$

So $\int xe^{-2x}$ $dx=-\frac{1}{2}xe^{-2x}+\frac{1}{2}\int e^{-2x}dx= -\frac{1}{2}xe^{-2x}-\frac{1}{4}e^{-2x}+C$

Solution:
Rewrite the integral as follows $\int \frac{2x}{e^{x}}\ dx=2\int xe^{-x}dx$

Then apply integration by parts method

$\int udv=uv-\int vdu$ with $u=x$
$dv=e^{-x}dx\Rightarrow du=dx$
$v=\int e^{-x}dx=-e^{-x}$ and now, we substitute in
$\int \frac{2x}{e^{x}}$ $dx=2\int xe^{-x}dx=2\left( -xe^{-x}-\int -e^{-x}dx\right) =-2xe^{-x}-2e^{-x}+C$

$\int \frac{2x}{e^{x}}$ $dx=-e^{-x}\left( 2x+2\right) +C$

Solution:
In this case, we apply twice the method of integration by parts, to reduce the degree of the power of $x$, from $2,1,0$
and simplify the integral

$u=x^{2}\qquad dv=e^{x}$ $dx$
then $du=2xdx\qquad v=\int e^{x}$ $dx=e^{x}$ and we replace in the formula
$\int udv=uv-\int vdu$ so,
$\int x^{2}e^{x}$ $dx=x^{2}e^{x}-2$ $\int xe^{x}dx$
We use integration by parts:
$\int xe^{x}dx$ with
$u=x\qquad dv=e^{x}$ $dx$
$u=dx\qquad v=\int e^{x}$ $dx=e^{x}$

then $\int x^{2}e^{x}$ $dx=x^{2}e^{x}-2\left( xe^{x}-\int e^{x}dx\right) =x^{2}e^{x}-2xe^{x}+2e^{x}+C$

So $\int x^{2}e^{x}$ $dx=e^{x}\left( x^{2}-2x+2\right) +C$

Solution:
Here the substitution is $u=\ln (t+1)\qquad dv=tdt$ so

$du=\frac{1}{t+1}dt\qquad v=\int tdt=\frac{1}{2}t^{2}$ and replacing in the formula

$\int udv=uv-\int vdu$ we obtain

$\int t\ \ln (t+1)\ dt=\frac{1}{2}t^{2}\ln (t+1)-\frac{1}{2}$

$\int \frac{t^{2}}{t+1}dt$ Now to solve this

Substitute $z=t+1$
then $t=z-1$ and $dz=dt$
So
$\int \frac{t^{2}}{t+1}dt=\int \frac{\left(z-1\right)^{2}}{z}dz$
$=\int \frac{z^{2}-2z+1}{z}dz=\int zdz-2\int dz+\int z^{-1}dz=\frac{z^{2}}{2}-2z+\ln (z)$
Since $z=t+1$

$\int \frac{t^{2}}{t+1}dt=\frac{\left( t+1\right)^{2}}{2}-2\left( t+1\right) +\ln (t+1)$ so $\int t\ln (t+1)dt=\frac{1}{2}t^{2}\ln (t+1)-\frac{1}{2}\left[ \frac{\left( t+1\right) ^{2}}{2}-2\left( t+1\right) +\ln (t+1)\right] +C$

$\int t$ $\ln (t+1)\ dt=\ln (t+1)\left[ \frac{1}{2}t^{2}-\frac{1}{2}\right] -\frac{1}{2}\left[ \frac{\left( t+1\right) ^{2}}{2}-2\left( t+1\right)\right] +C$

Solution:
In this case we do, $u=\ln (x)\qquad dv=x^{-2}dx$ then

$du=\frac{dx}{x}\qquad v=\int x^{-2}dx=-x^{-1}$ and now we replace in

$\int udv=uv-\int vdu$ then $\int \frac{\ln (x)}{x^{2}}dx=-\frac{\ln (x)}{x}+\int x^{-2}dx$ so

$\int \frac{\ln (x)}{x^{2}}dx=-\frac{\ln (x)}{x}-x^{-1}+C=-\frac{1}{x}\left( \ln x+1\right) +C$

Solution:
We have to do $u=x\qquad dv=\cos x$ $dx$ so $du=dx$ and

$v=\int \cos x\ dx=\sin x$ then if we apply the formula

$\int udv=uv-\int vdu$ we obtain
$\int x\cos x\ dx=x\sin x-\int \sin xdx$

So $\int x\cos x$ $dx=x\sin x+\cos x+C$

Solution:
Let $u=x\qquad dv=\sin xdx$ then $du=dx$ and $v=\int \sin xdx$
$v=-\cos x$, now we substitute in $\int udv=uv-\int vdu$ so
$\int x\sin x$ $dx=-x\cos x+\int \cos xdx=-x\cos x+\sin x+C$

Solution:
In this case, we must apply twice the method of integration
by parts, to reduce the degree of the power of $x$, from $2,1,0$
and simplify the integral, so we do,
$u=x^{2}\qquad dv=\sin x\ dx$
then $du=2xdx\qquad v=-\cos x$ then
$\int udv=uv-\int vdu$

$\int x^{2}\sin x$ $dx=-x^{2}\cos x-2\int x\cos xdx$
Now we apply integration by parts to the integral
$\int x\cos xdx=x\sin x-\int \sin xdx$

Let $u=x\qquad dv=\cos xdx$ and $du=dx\qquad v=\sin x$

So, $\int x\cos xdx=x\sin x+\cos x$ then $\int x^{2}\sin x\ dx=$
$=-x^{2}\cos x-2\left( x\sin x+\cos x\right) +C$

Solution:
Here we can find a recursive solution if $I=\int e^{2x}\sin x\ dx$
and we apply integration by parts, then $u=e^{2x},dv=\sin x\ dx$
$du=2e^{2x}dx\qquad v=-\cos x$
so $\int udv=uv-\int vdu\Longrightarrow \int e^{2x}\sin x\ dx=-e^{2x}\cos x+2\int \cos xe^{2x}dx$

now again we apply integration by parts,
so $u=e^{2x},dv=\cos x\ dx$
Then $du=2e^{2x}dx\qquad v=\sin x$
And replacing in the integral
$I=\int e^{2x}\sin x\ dx=-e^{2x}\cos x+2 \int \cos xe^{2x}dx$ now we apply integration by parts
$u=e^{2x}\qquad dv=\sin xdx$
$du=2e^{2x}dx\qquad v=-\cos x$

so, $I=\int e^{2x}\sin x\ dx=-e^{2x}\cos x+2\int -e^{2x}\cos xdx$
$=-e^{2x}\cos x-2\int e^{2x}\cos xdx=-e^{2x}\cos x-2I$ then
$3I=-e^{2x}\cos x+C\Rightarrow I=-\frac{1}{3}e^{2x}\cos x+C$
We get $I=\int e^{2x}\sin x\ dx=-\frac{1}{3}e^{2x}\cos x+C$

Solution:
$\int e^{-3x}\sin 5x$ $dx\qquad $ in this case we do
$u=e^{-3x}\qquad dv=\sin 5x\ dx$
$du=-\frac{1}{3}e^{-3x}dx\qquad v=-\frac{1}{5}\cos 5x$
Since $I=\int e^{-3x}\sin 5x\ dx$
$I=uv-vdu=-\frac{1}{5}e^{-3x}\cos 5x-\frac{1}{15}\int e^{-3x}\cos 5xdx$

We apply integration by parts to this integral again, so
$u=e^{-3x}\qquad dv=\cos 5x$ $dx\Longrightarrow du=-\frac{1}{3}e^{-3x}dx\qquad v=\frac{1}{5}\sin 5x$

Then $I=uv-vdu=-\frac{1}{5}e^{-3x}\cos 5x-\frac{1}{15}\int e^{-3x}\cos 5xdx$
So $I=-\frac{1}{5}e^{-3x}\cos 5x-\frac{1}{15}\left[ \frac{1}{5}e^{-3x}\sin 5x+\frac{1}{15}\int e^{-3x}\sin 5xdx\right] $

then $I+\frac{1}{225}I=-\frac{1}{5}e^{-3x}\cos 5x-\frac{1}{75}e^{-3x}\sin 5x$
and now
$\frac{226}{225}I=-\frac{1}{5}e^{-3x}\cos 5x-\frac{1}{75}e^{-3x}\sin 5x$ so
$I=\frac{225}{226}\left( -\frac{1}{5}e^{-3x}\cos 5x-\frac{1}{75}e^{-3x}\sin 5x\right) +C$

Solution:
If $u=\tan x\Rightarrow du=\sec ^{2}xdx$ and

$\int \sec ^{2}x\tan x\ dx=\int udu$

So $\int \sec ^{2}x\tan x$ $dx=\frac{u^{2}}{2}+C=\frac{\tan ^{2}x}{2}+C$

Solution:
We do $\int \tan ^{2}x\sec^{2}x\sec^{2}x\ dx$ and since
$\sec^{2}x=\tan ^{2}x+1$
We get $\int \tan^{2}x\sec^{4}x\ dx=\ \int \tan ^{2}x\left( \tan^{2}x+1\right) \sec ^{2}x\ dx$
$=\int \left( \tan ^{4}x+\tan^{2}x\right) \sec ^{2}x$ $dx$
We can do now:
If $u=\tan x\Rightarrow du=\sec^{2}xdx$ so $\int \tan^{2}x\sec^{4}x\ dx=$
$=\int \left( u^{4}+u^{2}\right) du=\frac{u^{5}}{5}+\frac{u^{3}}{3}+C$ then
$\int \tan^{2}x\sec^{4}x\ dx=\frac{\tan ^{5}x}{5}+\frac{\tan ^{3}x}{3}+C$