Integration by Trigonometric Substitution: Problems with Solutions

Solution:
Substitute $x=4\sin \theta$, so $dx=4\cos \theta d\theta $

Then $\sqrt{\left( 16-x^{2}\right) }=4\sqrt{1-\sin ^{2}\theta }=4\sqrt{\cos^{2}\theta }=4\cos \theta $

The given integral equals
$\int \frac{1}{\sqrt{\left( 16-x^{2}\right) }}dx=\int \frac{1}{4\cos \theta }4\cos \theta d\theta =\int d\theta $

$\theta =\arcsin \frac{x}{4}$ then $\int \frac{1}{\sqrt{\left(16-x^{2}\right) }}dx=\arcsin \frac{x}{4}+C$

Solution:
Substitute $x=5\sec \theta $ and $dx=5\sec \theta \tan \theta d\theta $

Then $x^{2}-25=25(\sec ^{2}\theta -1)=25\tan ^{2}\theta $ So
$\int \frac{1}{x^{2}-25}dx=\int \frac{1}{25\tan ^{2}\theta }5\sec \theta \tan \theta d\theta =\frac{1}{5}\int \frac{\sec \theta }{\tan \theta } d\theta =\frac{1}{5}\int \sec \theta \cot \theta d\theta $

$\int \frac{1}{x^{2}-25}dx=\frac{1}{5}\int \csc \theta d\theta $ $=-\ln \left\vert \csc \theta +\cot \theta \right\vert +C$

Convert back to x.
$\sec \theta =\frac{x}{5}\Longrightarrow \csc \theta =\frac{x}{\sqrt{25+x^{2}}}$ and

$\cot \theta =\frac{5}{\sqrt{25+x^{2}}}$ then $\int \frac{1}{x^{2}-25} dx=-\ln \left\vert \frac{x}{\sqrt{25+x^{2}}}+\frac{5}{\sqrt{25+x^{2}}} \right\vert +C$

Solution:
The integral may be evaluated by letting $x=\tan \theta $, $dx=\sec^{2}\theta d\theta $ so
$\int x\sqrt{1+x^{2}}dx=\int \tan \theta \sqrt{1+\tan \theta^{2}}\sec^{2}\theta d\theta =\int \tan \theta \sec^{3}\theta d\theta$

$\int \sec \theta \tan \theta \sec^{2}\theta d\theta$
If $u=\sec \theta, du=\sec \theta \tan \theta d\theta$

So $\int \sec \theta \tan \theta \sec^{2}\theta d\theta =\int u^{2}du=\frac{u^{3}}{3}+C=\frac{1}{3}\sec^{3}\theta +C$

We use the triangle.
So $\int x\sqrt{1+x^{2}}dx=\frac{1}{3}\left( 1+x^{2}\right) ^{3/2}+C$

Solution:
We may evaluate the integral by letting $x=\tan \theta $ $\ $then $dx=\sec ^{2}\theta d\theta $,

So $\int \frac{9x}{\sqrt{1+x^{2}}}dx=9\int \frac{\tan \theta }{\sqrt{1+\tan ^{2}}\theta }\sec ^{2}\theta d\theta =9\int \frac{\tan \theta }{\sec \theta }\sec ^{2}\theta d\theta =9\int \tan \theta \sec \theta d\theta =9\sec \theta +C$

Concidering the triangle below:
$\int \frac{9x}{\sqrt{1+x^{2}}}dx=9\sqrt{1+x^{2}}+C$

Solution:
We apply the method of completing the square in the expression

$4x-x^{2}=-\left( x-2\right)^{2}+4=4-\left( x-2\right)^{2}$
So $\int \frac{1}{\sqrt{4x-x^{2}}}dx=\int \frac{1}{\sqrt{4-\left( x-2\right)^{2}}}dx$.
Let $u=x-2\Longrightarrow du=dx$

Then $\int \frac{1}{\sqrt{4-\left( x-2\right)^{2}}}dx=\int \frac{1}{\sqrt{4-u^{2}}}du$ We make the trigonometric substitution $u=2\sin \theta$

$du=2\cos \theta d\theta \Longrightarrow \int \frac{1}{\sqrt{4-u^{2}}}du =\int \frac{1}{\sqrt{4-4\sin ^{2}\theta }}du$ $=\int \frac{2\cos \theta d\theta }{2\sqrt{\cos ^{2}\theta }}=\int d\theta =\theta +C=\arcsin \frac{u}{2}+C$
Convert back to x.
$u=x-2$

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

Solution:
Let $x=4\sin \theta \Longrightarrow dx=4\cos \theta d\theta $, then
$\int \frac{x^{2}}{\sqrt{16-x^{2}}}dx=\int \frac{4\sin ^{2}\theta }{\sqrt{ 16-16\sin ^{2}\theta }}4\cos \theta d\theta =\frac{16}{4}\int \frac{\sin^{2}\theta \cos \theta d\theta }{\sqrt{\cos ^{2}\theta }}$

$\int \frac{x^{2}}{\sqrt{16-x^{2}}}dx=4\int \sin^{2}\theta d\theta=2\theta -\sin 2\theta +C=2\theta -2\sin \theta \cos \theta +C$

Look at the triangle where $a=4$ and $\cos \theta =\frac{\sqrt{16-x^{2}}}{4}$

So $\int \frac{x^{2}}{\sqrt{16-x^{2}}}dx=2\arcsin \frac{x}{4}-\frac{x}{2}\frac{\sqrt{16-x^{2}}}{4}+C$

Solution:
After completing the square we get:
$x^{2}+6x+12=\left( x+3\right)^{2}+3\Longrightarrow \int \frac{x}{\sqrt{ x^{2}+6x+12}}dx=\int \frac{x}{\sqrt{\left( x+3\right) ^{2}+3}}dx$

Make the substitution
$u=x+3\Longrightarrow du=dx,x=u-3$ So
$\int \frac{x}{\sqrt{\left( x+3\right) ^{2}+3}}dx=\int \frac{\left( u-3\right) }{\sqrt{u^{2}+3}}du$

Then make the trigonometric substitution $u=\sqrt{3}\tan \theta $
$du=\sqrt{3}\sec ^{2}\theta d\theta $

$\int \frac{\left( u-3\right) }{\sqrt{u^{2}+3}}du=\int \frac{\left( \sqrt{3} \tan \theta -3\right) }{\sqrt{3\left( \tan ^{2}\theta +1\right) }}\sqrt{3} \sec ^{2}\theta d\theta =\int \frac{\left( \sqrt{3}\tan \theta -3\right) }{\sqrt{\sec ^{2}\theta }} \sec ^{2}\theta d\theta =\sqrt{3}\int \sec \theta \tan \theta d\theta -3\int \sec \theta d\theta =\sqrt{3}\sec \theta -3\ln \left\vert \sec \theta +\tan \theta \right\vert +C$

Take a look at the triangle below where $a=\sqrt{3},u=x$

So $=\sqrt{3}\frac{\sqrt{3+u^{2}}}{\sqrt{3}}-3\ln \left\vert \frac{\sqrt{ 3+u^{2}}}{\sqrt{3}}+\frac{u}{\sqrt{3}}\right\vert +C$

Convert back to x
$u=x+3$

$\int \frac{x}{\sqrt{x^{2}+6x+12}}dx=\sqrt{3}\frac{\sqrt{3+\left( x+3\right)^{2}}}{\sqrt{3}}-3\ln \left\vert \frac{\sqrt{3+\left( x+3\right) ^{2}}}{\sqrt{3}}+\frac{\left( x+3\right) }{\sqrt{3}}\right\vert +C$

Solution:
Since $2x-x^{2}=-(x-1)^{2}+1$,
$\int \frac{x^{2}}{\sqrt{2x-x^{2}}}dx=\int \frac{x^{2}}{\sqrt{1-(x-1)^{2}}}dx$

Let $u=x-1\Longrightarrow du=dx$ $x=u+1$ then $\int \frac{x^{2}}{\sqrt{1-(x-1)^{2}}}dx=\int \frac{\left( u+1\right) ^{2}}{\sqrt{1-u^{2}}}du$
We apply trigonometric substitution.
So, $u=\sin \theta \Longrightarrow du=\cos \theta d\theta $ and $\sqrt{1-u^{2}}=\sqrt{\cos ^{2}\theta }=\cos \theta $ so, $\int \frac{\left( u+1\right) ^{2}}{\sqrt{1-u^{2}}}du=\int \frac{\left( \sin \theta +1\right)^{2}}{\cos \theta }\cos \theta d\theta=\int \left( \sin ^{2}\theta +2\sin \theta \cos \theta +1\right) d\theta =\int \sin ^{2}\theta d\theta +2\int \sin \theta \cos \theta d\theta +\int d\theta =\frac{\theta }{2}-\frac{1}{4}\sin 2\theta +2\frac{\sin ^{2}\theta }{2} +\theta +C\ =\frac{3\theta }{2}-\frac{1}{2}\sin \theta \cos \theta +\sin^{2}\theta +C$

Considering the triangle below where $x=u,a=1$ and $\theta =\arcsin u$ we get
$u=\sin \theta \qquad \cos \theta =\sqrt{1-u^{2}}$ then

$\int \frac{\left( u+1\right) ^{2}}{\sqrt{1-u^{2}}}du=\frac{3}{2}\arcsin u-\frac{1}{2}u\sqrt{1-u^{2}}+u^{2}+C$

Since $u=x-1\Longrightarrow \int \frac{x^{2}}{\sqrt{2x-x^{2}}}dx=\frac{3}{2}\arcsin \left( x-1\right) -\frac{1}{2}\left( x-1\right) \sqrt{1-\left( x-1\right) ^{2}}+\left( x-1\right)^{2}+C$

Solution:
Answer: Only (i) (iii) are true.

$\int \frac{x-9}{x^{2}-6x}dx\Longrightarrow $(i) $\int \frac{x-9}{x\left( x-6\right) }dx$

(iii) $\frac{x-9}{x\left( x-6\right) }=\frac{A}{x}+\frac{B}{x-6}=\frac{ A\left( x-6\right) +Bx}{x\left( x-6\right) }$
We get $x-9=A\left( x-6\right) +Bx$

Then $x-9=\left( A+B\right) x-6A\Longrightarrow \left\{ \begin{array}{c} A+B=1 \\ -6A=-9 \end{array} \right\} \Longrightarrow A=\frac{3}{2},B=-\frac{1}{2}$

Then $\int \frac{x-9}{x\left( x-6\right) }dx=\int \left( \frac{\frac{3}{2}}{x }-\frac{\frac{1}{2}}{x-6}\right) dx=\frac{3}{2}\int \frac{1}{x}dx-\frac{1}{2} \int \frac{1}{x-6}dx=\frac{3}{2}\ln x-\frac{1}{2}\ln \left\vert x-6\right\vert +C=\ln \sqrt{ x^{3}}-\ln \sqrt{x-6}+\ln C=\ln \left\vert C\sqrt{\frac{x^{3}}{x-6}}\right\vert $

Solution:
$\int \frac{1}{x^{2}-9}dx=\int \frac{1}{\left( x+3\right) \left( x-3\right) } dx$
We obtain the partial fractions

$\frac{1}{\left( x+3\right) \left( x-3\right) }=\frac{A}{x+3}+\frac{B}{x-3}= \frac{A\left( x-3\right) +B\left( x+3\right) }{\left( x+3\right) \left(x-3\right) }\Longrightarrow $

$1=\left( A+B\right) x+3\left( B-A\right)$

So we get $\left\{ \begin{array}{c} A+B=0 \\ 3\left( B-A\right) =1 \end{array} \right\}$

Since $A=-B\Longrightarrow 3\left( B-A\right) =3\left( B+B\right)=1\Longrightarrow B=\frac{1}{6}$

And $A=-\frac{1}{6}$
So $\int \frac{1}{x^{2}-9}dx=\int \left( \frac{-\frac{1}{6}}{x+3}+\frac{\frac{1}{6}}{x-3}\right) dx=-\frac{1}{6}\int \frac{1}{x+3}dx+\frac{1}{6}\int \frac{1}{x-3}dx=\frac{1}{6}(\ln \left( x-3\right) -\ln (x+3))+C$

$\int \frac{1}{x^{2}-9}dx=\ln (C\sqrt[6]{\frac{x-3}{x+3}})$

Solution:
We must factor the denominator $x^{2}+3x-4=\left( x+4\right) \left(x-1\right)$

Then $\frac{5}{x^{2}+3x-4}=\frac{A}{x+4}+\frac{B}{x-1}=\frac{A(x-1)+B(x+4)}{x^{2}+3x-4}$
So
$(A+B)x+(4B-A)=5\Longrightarrow \left\{ \begin{array}{c} A+B=0 \\ 4B-A=5 \end{array} \right\}$
Since $A=-B$
Then $4B-A=5B=5\Longrightarrow B=1,A=-1$ so

$\int \frac{5}{x^{2}+3x-4}dx=\int \left( \frac{-1}{x+4}+\frac{1}{x-1}\right)dx=\ln \left\vert x-1\right\vert -\ln \left\vert x+1\right\vert +\ln C$

Then $\int \frac{5}{x^{2}+3x-4}dx=\ln \left\vert C\frac{x-1}{x+1}\right\vert$

Solution:
Since $x^{2}+11x+18=\left( x+9\right) \left( x+2\right) $

Then $\frac{x+2}{x^{2}+11x+18}=\frac{A}{x+9}+\frac{B}{x+2}=\frac{A(x+2)+B(x+9)}{x^{2}+11x+18}\Longrightarrow $

$x+2=\left( A+B\right) x+(2A+9B)\Longrightarrow \left\{ \begin{array}{c} A+B=1 \\ 2A+9B=2 \end{array} \right\} $

Then $A=-B$ and $2A-9A=2\Longrightarrow A=-\frac{2}{7},B=\frac{2}{7}$

So $\int \frac{x+2}{x^{2}+11x+18}dx=\int \left( \frac{-\frac{2}{7}}{x+9}+\frac{\frac{2}{7}}{x+2}\right) dx=\frac{2}{7}\int \frac{1}{x+2}dx-\frac{2}{7}\int \frac{1}{x+9}dx$

$\ln \sqrt[7]{\left( x+2\right) ^{2}}-\ln \sqrt[7]{\left( x+9\right)^{2}}+\ln C=\ln \left\vert C\sqrt[7]{\left( \frac{x+9}{x+2}\right)^{2}} \right\vert $

Solution:
Answer: Perform (i) and (iii) only.

$\int \frac{5x^{2}-12x-12}{x^{3}-4x}dx\Longrightarrow \frac{5x^{2}-12x-12}{x^{3}-4x}=\frac{5(x^{2}-\frac{12}{5}x-\frac{12}{5})}{x\left( x+2\right)\left( x-2\right) }=\frac{5(x-\frac{6}{5}-\frac{4}{5}\sqrt{6})(x-\frac{6}{5}+\frac{4}{5}\sqrt{6})}{x\left( x+2\right) \left( x-2\right) }$ (i) now $\frac{5(x-\frac{6}{5}- \frac{4}{5}\sqrt{6})(x-\frac{6}{5}+\frac{4}{5}\sqrt{6})}{x\left( x+2\right)\left( x-2\right) }=\frac{A}{x}+\frac{B}{x+2}+\frac{C}{x-2}=\frac{A(x+2)(x-2)+Bx(x-2)+Cx(x+2)}{ x\left( x+2\right) \left( x-2\right) }$

Then $A(x+2)(x-2)+Bx(x-2)+Cx(x+2)=5(x-\frac{6}{5}-\frac{4}{5}\sqrt{6})(x-\frac{6}{5}+\frac{4}{5}\sqrt{6})$

If $x=2$ then $8C=5(2-\frac{6}{5}-\frac{4}{5}\sqrt{6})(2-\frac{6}{5}+\frac{4}{5}\sqrt{6})= -16$

So $C=-2$ if $x=-2$ then $8B=5(-2-\frac{6}{5}-\frac{4}{5}\sqrt{6})(-2-\frac{6}{5}+\frac{4}{5}\sqrt{6})$
$8B=32$ so $B=4$ and if $x=0$ then
$-4A=5(-\frac{6}{5}-\frac{4}{5}\sqrt{6})(-\frac{6}{5}+\frac{4}{5}\sqrt{6} )= -12$ so $A=3$

$\int \frac{5x^{2}-12x-12}{x^{3}-4x}dx=\int \left( \frac{3}{x}+\frac{4}{x+2}-\frac{2}{x-2}\right) dx=3\ln x+4\ln \left\vert x+2\right\vert -2\ln \left\vert x-2\right\vert +\ln C=\ln \left\vert C\frac{x^{3}\left( x+2\right)^{4}}{\left( x-2\right) ^{2}}\right\vert $

Solution:
Answer: Perform (i) and (iii) only.
$\frac{x+2}{x^{2}-4x}=\frac{x+2}{x(x-4)}=\frac{A}{x}+\frac{B}{x-4}=\frac{ A(x-4)+Bx}{x(x-4)}$ then $A(x-4)+Bx=x+2$ so $\left( A+B\right) x-4A=x+2\Longrightarrow \left\{ \begin{array}{c} A+B=1 \\ -4A=2 \end{array} \right\} $ then $A=-\frac{1}{2}$ $B=\frac{3}{2}$ then $\int \frac{x+2}{x^{2}-4x}dx=\int \left( \frac{-\frac{1 }{2}}{x}+\frac{\frac{3}{2}}{x-4}\right) dx=-\frac{1}{2}\int \frac{1}{x}dx+ \frac{3}{2}\int \frac{1}{x-4}dx$ $=-\frac{1}{2}\ln x+\frac{3}{2}\ln \left\vert x-4\right\vert +\ln C=\ln \left\vert C\sqrt{\frac{\left( x-4\right) ^{3}}{x}}\right\vert $

Solution:
Answer: Perform (i), (ii) and (iii)

$\frac{x^{2}+2x+1}{x(x-1)(x+1)}=\frac{\left( x+1\right)^{2}}{x(x-1)(x+1)}= \frac{x+1}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1}=\frac{A(x-1)+Bx}{x(x-1)}$

Then $x+1=\left( A+B\right) x-A\Longrightarrow \left\{ \begin{array}{c} \left( A+B\right) =1 \\ -A=1 \end{array} \right\} \Longrightarrow A=-1$

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