Examples (Revised) - Chapter 7 - Integrals - Ncert Solutions class 12 - Maths
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Example 1
Write an anti derivative for each of the following functions using the method of inspection:
(i) $\cos 2 x$
(ii) $3 x^2+4 x^3$
(iii) $\frac{1}{x}, x \neq 0$
Solution
(i) We look for a function whose derivative is $\cos 2 x$. Recall that
$
\frac{d}{d x} \sin 2 x=2 \cos 2 x
$
or $\quad \cos 2 x=\frac{1}{2} \frac{d}{d x}(\sin 2 x)=\frac{d}{d x}\left(\frac{1}{2} \sin 2 x\right)$
Therefore, an anti derivative of $\cos 2 x$ is $\frac{1}{2} \sin 2 x$.
(ii) We look for a function whose derivative is $3 x^2+4 x^3$. Note that
$
\frac{d}{d x}\left(x^3+x^4\right)=3 x^2+4 x^3
$
Therefore, an anti derivative of $3 x^2+4 x^3$ is $x^3+x^4$.
(iii) We know that
$
\frac{d}{d x}(\log x)=\frac{1}{x}, x>0 \text { and } \frac{d}{d x}[\log (-x)]=\frac{1}{-x}(-1)=\frac{1}{x}, x<0
$
Combining above, we get $\frac{d}{d x}(\log |x|)=\frac{1}{x}, x \neq 0$
Therefore, $\int \frac{1}{x} d x=\log |x|$ is one of the anti derivatives of $\frac{1}{x}$.
Example 2
Find the following integrals:
(i) $\int \frac{x^3-1}{x^2} d x$
(ii) $\int\left(x^{\frac{2}{3}}+1\right) d x$
(iii) $\int\left(x^{\frac{3}{2}}+2 e^x-\frac{1}{x}\right) d x$
Solution
(i) We have
$
\begin{aligned}
\int \frac{x^3-1}{x^2} & d x=\int x d x-\int x^{-2} d x \quad \text { (by Property V) } \\
& =\left(\frac{x^{1+1}}{1+1}+\mathrm{C}_1\right)-\left(\frac{x^{-2+1}}{-2+1}+\mathrm{C}_2\right) ; \mathrm{C}_1, \mathrm{C}_2 \text { are constants of integration } \\
& =\frac{x^2}{2}+\mathrm{C}_1-\frac{x^{-1}}{-1}-\mathrm{C}_2=\frac{x^2}{2}+\frac{1}{x}+\mathrm{C}_1-\mathrm{C}_2 \\
& =\frac{x^2}{2}+\frac{1}{x}+\mathrm{C}, \text { where } \mathrm{C}=\mathrm{C}_1-\mathrm{C}_2 \text { is another constant of integration. }
\end{aligned}
$
(ii) We have
$
\begin{aligned}
\int\left(x^{\frac{2}{3}}+1\right) d x & =\int x^{\frac{2}{3}} d x+\int d x \\
& =\frac{x^{\frac{2}{3}+1}}{\frac{2}{3}+1}+x+\mathrm{C}=\frac{3}{5} x^{\frac{5}{3}}+x+\mathrm{C}
\end{aligned}
$
(iii) We have $\int\left(x^{\frac{3}{2}}+2 e^x-\frac{1}{x}\right) d x=\int x^{\frac{3}{2}} d x+\int 2 e^x d x-\int \frac{1}{x} d x$
$
\begin{aligned}
& =\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}+2 e^x-\log |x|+\mathrm{C} \\
& =\frac{2}{5} x^{\frac{5}{2}}+2 e^x-\log |x|+\mathrm{C}
\end{aligned}
$
Example 3
Find the following integrals:
(i) $\int(\sin x+\cos x) d x$
(ii) $\int \operatorname{cosec} x(\operatorname{cosec} x+\cot x) d x$
(iii) $\int \frac{1-\sin x}{\cos ^2 x} d x$
Solution
(i) $\mathrm{We}$ have
$
\begin{aligned}
\int(\sin x+\cos x) d x & =\int \sin x d x+\int \cos x d x \\
& =-\cos x+\sin x+C
\end{aligned}
$
(ii) We have
$
\begin{aligned}
\int(\operatorname{cosec} x(\operatorname{cosec} x+\cot x) d x & =\int \operatorname{cosec}^2 x d x+\int \operatorname{cosec} x \cot x d x \\
& =-\cot x-\operatorname{cosec} x+\mathrm{C}
\end{aligned}
$
(iii) We have
$
\begin{aligned}
\int \frac{1-\sin x}{\cos ^2 x} d x & =\int \frac{1}{\cos ^2 x} d x-\int \frac{\sin x}{\cos ^2 x} d x \\
& =\int \sec ^2 x d x-\int \tan x \sec x d x \\
& =\tan x-\sec x+C
\end{aligned}
$
Example 4
Find the anti derivative $\mathrm{F}$ of $f$ defined by $f(x)=4 x^3-6$, where $\mathrm{F}(0)=3$
Solution
One anti derivative of $f(x)$ is $x^4-6 x$ since
$
\frac{d}{d x}\left(x^4-6 x\right)=4 x^3-6
$
Therefore, the anti derivative $\mathrm{F}$ is given by
$
\mathrm{F}(x)=x^4-6 x+\mathrm{C} \text {, where } \mathrm{C} \text { is constant. }
$
Given that
$
\begin{aligned}
\mathrm{F}(0) & =3, \text { which gives, } \\
3 & =0-6 \times 0+\mathrm{C} \text { or } \mathrm{C}=3
\end{aligned}
$
Hence, the required anti derivative is the unique function $\mathrm{F}$ defined by
$
\mathrm{F}(x)=x^4-6 x+3 .
$
Example 5
Integrate the following functions w.r.t. $x$ :
(i) $\sin m x$
(ii) $2 x \sin \left(x^2+1\right)$
(iii) $\frac{\tan ^4 \sqrt{x} \sec ^2 \sqrt{x}}{\sqrt{x}}$
(iv) $\frac{\sin \left(\tan ^{-1} x\right)}{1+x^2}$
Solution
(i) We know that derivative of $m x$ is $m$. Thus, we make the substitution $m x=t$ so that $m d x=d t$.
Therefore, $\quad \int \sin m x d x=\frac{1}{m} \int \sin t d t=-\frac{1}{m} \cos t+\mathrm{C}=-\frac{1}{m} \cos m x+\mathrm{C}$
(ii) Derivative of $x^2+1$ is $2 x$. Thus, we use the substitution $x^2+1=t$ so that $2 x d x=d t$
Therefore, $\int 2 x \sin \left(x^2+1\right) d x=\int \sin t d t=-\cos t+\mathrm{C}=-\cos \left(x^2+1\right)+\mathrm{C}$
(iii) Derivative of $\sqrt{x}$ is $\frac{1}{2} x^{-\frac{1}{2}}=\frac{1}{2 \sqrt{x}}$. Thus, we use the substitution
$\sqrt{x}=t$ so that $\frac{1}{2 \sqrt{x}} d x=d t$ giving $d x=2 t d t$.
Thus, $\quad \int \frac{\tan ^4 \sqrt{x} \sec ^2 \sqrt{x}}{\sqrt{x}} d x=\int \frac{2 t \tan ^4 t \sec ^2 t d t}{t}=2 \int \tan ^4 t \sec ^2 t d t$
Again, we make another substitution $\tan t=u$ so that $\sec ^2 t d t=d u$
Therefore,
$
\begin{aligned}
2 \int \tan ^4 t \sec ^2 t d t & =2 \int u^4 d u=2 \frac{u^5}{5}+\mathrm{C} \\
& =\frac{2}{5} \tan ^5 t+\mathrm{C}(\text { since } u=\tan t) \\
& =\frac{2}{5} \tan ^5 \sqrt{x}+\mathrm{C}(\text { since } t=\sqrt{x})
\end{aligned}
$
Hence, $\quad \quad \quad \quad \frac{\tan ^4 \sqrt{x} \sec ^2 \sqrt{x}}{\sqrt{x}} d x=\frac{2}{5} \tan ^5 \sqrt{x}+\mathrm{C}$
Alternatively, make the substitution $\tan \sqrt{x}=t$
(iv) Derivative of $\tan ^{-1} x=\frac{1}{1+x^2}$. Thus, we use the substitution
$
\tan ^{-1} x=t \text { so that } \frac{d x}{1+x^2}=d t \text {. }
$
Therefore, $\int \frac{\sin \left(\tan ^{-1} x\right)}{1+x^2} d x=\int \sin t d t=-\cos t+\mathrm{C}=-\cos \left(\tan ^{-1} x\right)+\mathrm{C}$
Now, we discuss some important integrals involving trigonometric functions and their standard integrals using substitution technique. These will be used later without reference.
(i) $\int \tan x d x=\log |\sec x|+C$
We have
$
\int \tan x d x=\int \frac{\sin x}{\cos x} d x
$
Put $\cos x=t$ so that $\sin x d x=-d t$
Then
$
\int \tan x d x=-\int \frac{d t}{t}=-\log |t|+\mathrm{C}=-\log |\cos x|+\mathrm{C}
$
$
\int \tan x d x=\log |\sec x|+\mathrm{C}
$
(ii) $\int \cot x d x=\log |\sin x|+C$
We have $\int \cot x d x=\int \frac{\cos x}{\sin x} d x$
Put $\sin x=t$ so that $\cos x d x=d t$
Then
$
\int \cot x d x=\int \frac{d t}{t}=\log |t|+\mathrm{C}=\log |\sin x|+\mathrm{C}
$
(iii) $\int \sec x d x=\log |\sec x+\tan x|+C$
We have
$
\int \sec x d x=\int \frac{\sec x(\sec x+\tan x)}{\sec x+\tan x} d x
$
Put $\sec x+\tan x=t$ so that $\sec x(\tan x+\sec x) d x=d t$
Therefore, $\int \sec x d x=\int \frac{d t}{t}=\log |t|+\mathrm{C}=\log |\sec x+\tan x|+\mathrm{C}$
(iv) $\int \operatorname{cosec} x d x=\log |\operatorname{cosec} x-\cot x|+C$
We have
$
\int \operatorname{cosec} x d x=\int \frac{\operatorname{cosec} x(\operatorname{cosec} x+\cot x)}{(\operatorname{cosec} x+\cot x)} d x
$
Put $\operatorname{cosec} x+\cot x=t$ so that $-\operatorname{cosec} x(\operatorname{cosec} x+\cot x) d x=d t$
So
$
\begin{aligned}
\int \operatorname{cosec} x d x & =-\int \frac{d t}{t}=-\log |t|=-\log |\operatorname{cosec} x+\cot x|+\mathrm{C} \\
& =-\log \left|\frac{\operatorname{cosec}^2 x-\cot ^2 x}{\operatorname{cosec} x-\cot x}\right|+\mathrm{C} \\
& =\log |\operatorname{cosec} x-\cot x|+\mathrm{C}
\end{aligned}
$
Example 6
Find the following integrals:
(i) $\int \sin ^3 x \cos ^2 x d x$
(ii) $\int \frac{\sin x}{\sin (x+a)} d x$
(iii) $\int \frac{1}{1+\tan x} d x$
Solution
(i) We have
$
\begin{aligned}
\int \sin ^3 x \cos ^2 x d x & =\int \sin ^2 x \cos ^2 x(\sin x) d x \\
& =\int\left(1-\cos ^2 x\right) \cos ^2 x(\sin x) d x
\end{aligned}
$
Put $t=\cos x$ so that $d t=-\sin x d x$
Therefore, $\quad \int \sin ^2 x \cos ^2 x(\sin x) d x=-\int\left(1-t^2\right) t^2 d t$
$
\begin{aligned}
& =-\int\left(t^2-t^4\right) d t=-\left(\frac{t^3}{3}-\frac{t^5}{5}\right)+\mathrm{C} \\
& =-\frac{1}{3} \cos ^3 x+\frac{1}{5} \cos ^5 x+\mathrm{C}
\end{aligned}
$
(ii) Put $x+a=t$. Then $d x=d t$. Therefore
$
\begin{aligned}
\int \frac{\sin x}{\sin (x+a)} d x & =\int \frac{\sin (t-a)}{\sin t} d t \\
& =\int \frac{\sin t \cos a-\cos t \sin a}{\sin t} d t \\
& =\cos a \int d t-\sin a \int \cot t d t \\
& =(\cos a) t-(\sin a)\left[\log |\sin t|+\mathrm{C}_1\right] \\
& =(\cos a)(x+a)-(\sin a)\left[\log |\sin (x+a)|+\mathrm{C}_1\right] \\
& =x \cos a+a \cos a-(\sin a) \log |\sin (x+a)|-\mathrm{C}_1 \sin a
\end{aligned}
$
Hence, $\int \frac{\sin x}{\sin (x+a)} d x=x \cos a-\sin a \log |\sin (x+a)|+\mathrm{C}$,
where, $\mathrm{C}=-\mathrm{C}_1 \sin a+a \cos a$, is another arbitrary constant.
(iii)
$
\begin{aligned}
\int \frac{d x}{1+\tan x} & =\int \frac{\cos x d x}{\cos x+\sin x} \\
& =\frac{1}{2} \int \frac{(\cos x+\sin x+\cos x-\sin x) d x}{\cos x+\sin x} \\
& =\frac{1}{2} \int d x+\frac{1}{2} \int \frac{\cos x-\sin x}{\cos x+\sin x} d x \\
& =\frac{x}{2}+\frac{C_1}{2}+\frac{1}{2} \int \frac{\cos x-\sin x}{\cos x+\sin x} d x
\end{aligned}
$
Now, consider $\mathrm{I}=\int \frac{\cos x-\sin x}{\cos x+\sin x} d x$
Put $\cos x+\sin x=t$ so that $(\cos x-\sin x) d x=d t$
Therefore $\quad \mathrm{I}=\int \frac{d t}{t}=\log |t|+\mathrm{C}_2=\log |\cos x+\sin x|+\mathrm{C}_2$
Putting it in (1), we get
$
\begin{aligned}
\int \frac{d x}{1+\tan x} & =\frac{x}{2}+\frac{\mathrm{C}_1}{2}+\frac{1}{2} \log |\cos x+\sin x|+\frac{\mathrm{C}_2}{2} \\
& =\frac{x}{2}+\frac{1}{2} \log |\cos x+\sin x|+\frac{\mathrm{C}_1}{2}+\frac{\mathrm{C}_2}{2} \\
& =\frac{x}{2}+\frac{1}{2} \log |\cos x+\sin x|+C,\left(\mathrm{C}=\frac{\mathrm{C}_1}{2}+\frac{\mathrm{C}_2}{2}\right)
\end{aligned}
$
Example 7
Find (i) $\int \cos ^2 x d x$
(ii) $\int \sin 2 x \cos 3 x d x$
(iii) $\int \sin ^3 x d x$
Solution
(i) Recall the identity $\cos 2 x=2 \cos ^2 x-1$, which gives
$
\cos ^2 x=\frac{1+\cos 2 x}{2}
$
Therefore, $\quad \int \cos ^2 x d x=\frac{1}{2} \int(1+\cos 2 x) d x=\frac{1}{2} \int d x+\frac{1}{2} \int \cos 2 x d x$
$
=\frac{x}{2}+\frac{1}{4} \sin 2 x+\mathrm{C}
$
(ii) Recall the identity $\sin x \cos y=\frac{1}{2}[\sin (x+y)+\sin (x-y)]$
(Why?)
Then $\int \sin 2 x \cos 3 x d x=\frac{1}{2}\left[\int \sin 5 x d x-\int \sin x d x\right]$
$
\begin{aligned}
& =\frac{1}{2}\left[-\frac{1}{5} \cos 5 x+\cos x\right]+C \\
& =-\frac{1}{10} \cos 5 x+\frac{1}{2} \cos x+C
\end{aligned}
$
(iii) From the identity $\sin 3 x=3 \sin x-4 \sin ^3 x$, we find that
$
\sin ^3 x=\frac{3 \sin x-\sin 3 x}{4}
$
Therefore, $\quad \int \sin ^3 x d x=\frac{3}{4} \int \sin x d x-\frac{1}{4} \int \sin 3 x d x$
$
=-\frac{3}{4} \cos x+\frac{1}{12} \cos 3 x+C
$
Alternatively, $\int \sin ^3 x d x=\int \sin ^2 x \sin x d x=\int\left(1-\cos ^2 x\right) \sin x d x$
Put $\cos x=t$ so that $-\sin x d x=d t$
Therefore,
$
\begin{aligned}
\int \sin ^3 x d x & =-\int\left(1-t^2\right) d t=-\int d t+\int t^2 d t=-t+\frac{t^3}{3}+\mathrm{C} \\
& =-\cos x+\frac{1}{3} \cos ^3 x+\mathrm{C}
\end{aligned}
$
Example 8
Find the following integrals:
(i) $\int \frac{d x}{x^2-16}$
(ii) $\int \frac{d x}{\sqrt{2 x-x^2}}$
Solution
(i) We have $\int \frac{d x}{x^2-16}=\int \frac{d x}{x^2-4^2}=\frac{1}{8} \log \left|\frac{x-4}{x+4}\right|+$ C $[$ by 7.4 (1) $]$
(ii) $\int \frac{d x}{\sqrt{2 x-x^2}}=\int \frac{d x}{\sqrt{1-(x-1)^2}}$
Put $x-1=t$. Then $d x=d t$.
Therefore,
$
\begin{aligned}
\int \frac{d x}{\sqrt{2 x-x^2}} & =\int \frac{d t}{\sqrt{1-t^2}}=\sin ^{-1}(t)+\mathrm{C} \\
& =\sin ^{-1}(x-1)+\mathrm{C}
\end{aligned}
$
Example 9
Find the following integrals :
(i) $\int \frac{d x}{x^2-6 x+13}$
(ii) $\int \frac{d x}{3 x^2+13 x-10}$
(iii) $\int \frac{d x}{\sqrt{5 x^2-2 x}}$
Solution
(i) We have $x^2-6 x+13=x^2-6 x+3^2-3^2+13=(x-3)^2+4$
So,
$
\int \frac{d x}{x^2-6 x+13}=\int \frac{1}{(x-3)^2+2^2} d x
$
Let
$
x-3=t \text {. Then } d x=d t
$
Therefore
$
\begin{aligned}
\int \frac{d x}{x^2-6 x+13} & =\int \frac{d t}{t^2+2^2}=\frac{1}{2} \tan ^{-1} \frac{t}{2}+\mathrm{C} \\
& =\frac{1}{2} \tan ^{-1} \frac{x-3}{2}+\mathrm{C}
\end{aligned}
$
(ii) The given integral is of the form 7.4 (7). We write the denominator of the integrand,
$
\begin{aligned}
3 x^2+13 x-10 & =3\left(x^2+\frac{13 x}{3}-\frac{10}{3}\right) \\
& =3\left[\left(x+\frac{13}{6}\right)^2-\left(\frac{17}{6}\right)^2\right] \text { (completing the square) }
\end{aligned}
$
Thus $\int \frac{d x}{3 x^2+13 x-10}=\frac{1}{3} \int \frac{d x}{\left(x+\frac{13}{6}\right)^2-\left(\frac{17}{6}\right)^2}$
Put $x+\frac{13}{6}=t$. Then $d x=d t$.
Therefore, $\quad \int \frac{d x}{3 x^2+13 x-10}=\frac{1}{3} \int \frac{d t}{t^2-\left(\frac{17}{6}\right)^2}$
$
=\frac{1}{3 \times 2 \times \frac{17}{6}} \log \left|\frac{t-\frac{17}{6}}{t+\frac{17}{6}}\right|+\mathrm{C}_1
$
$[$ by $7.4(\mathrm{i})]$
$\begin{aligned}
& =\frac{1}{17} \log \left|\frac{x+\frac{13}{6}-\frac{17}{6}}{x+\frac{13}{6}+\frac{17}{6}}\right|+\mathrm{C}_1 \\
& =\frac{1}{17} \log \left|\frac{6 x-4}{6 x+30}\right|+\mathrm{C}_1 \\
& =\frac{1}{17} \log \left|\frac{3 x-2}{x+5}\right|+\mathrm{C}_1+\frac{1}{17} \log \frac{1}{3} \\
& =\frac{1}{17} \log \left|\frac{3 x-2}{x+5}\right|+\mathrm{C}, \text { where } \mathrm{C}=\mathrm{C}_1+\frac{1}{17} \log \frac{1}{3}
\end{aligned}$
(iii) We have
$
\text { have } \begin{aligned}
\int \frac{d x}{\sqrt{5 x^2-2 x}} & =\int \frac{d x}{\sqrt{5\left(x^2-\frac{2 x}{5}\right)}} \\
& =\frac{1}{\sqrt{5}} \int \frac{d x}{\sqrt{\left(x-\frac{1}{5}\right)^2-\left(\frac{1}{5}\right)^2}} \text { (completing the square) }
\end{aligned}
$
Put $x-\frac{1}{5}=t$. Then $d x=d t$.
Therefore,
$
\begin{aligned}
\int \frac{d x}{\sqrt{5 x^2-2 x}} & =\frac{1}{\sqrt{5}} \int \frac{d t}{\sqrt{t^2-\left(\frac{1}{5}\right)^2}} \\
& =\frac{1}{\sqrt{5}} \log \left|t+\sqrt{t^2-\left(\frac{1}{5}\right)^2}\right|+\mathrm{C} \\
& =\frac{1}{\sqrt{5}} \log \left|x-\frac{1}{5}+\sqrt{x^2-\frac{2 x}{5}}\right|+\mathrm{C}
\end{aligned}
$
[by 7.4 (4)]
Example 10
Find the following integrals:
(i) $\int \frac{x+2}{2 x^2+6 x+5} d x$
(ii) $\int \frac{x+3}{\sqrt{5-4 x-x^2}} d x$
Solution
(i) Using the formula 7.4 (9), we express
$
x+2=\mathrm{A} \frac{d}{d x}\left(2 x^2+6 x+5\right)+\mathrm{B}=\mathrm{A}(4 x+6)+\mathrm{B}
$
Equating the coefficients of $x$ and the constant terms from both sides, we get $4 \mathrm{~A}=1$ and $6 \mathrm{~A}+\mathrm{B}=2$ or $\mathrm{A}=\frac{1}{4}$ and $\mathrm{B}=\frac{1}{2}$.
Therefore, $\quad \int \frac{x+2}{2 x^2+6 x+5}=\frac{1}{4} \int \frac{4 x+6}{2 x^2+6 x+5} d x+\frac{1}{2} \int \frac{d x}{2 x^2+6 x+5}$
$
=\frac{1}{4} \mathrm{I}_1+\frac{1}{2} \mathrm{I}_2 \quad(\text { say })
$
In $\mathrm{I}_1$, put $2 x^2+6 x+5=t$, so that $(4 x+6) d x=d t$
Therefore,
$
\begin{aligned}
\mathrm{I}_1 & =\int \frac{d t}{t}=\log |t|+\mathrm{C}_1 \\
& =\log \left|2 x^2+6 x+5\right|+C_1
\end{aligned}
$
and
$
\begin{aligned}
I_2 & =\int \frac{d x}{2 x^2+6 x+5}=\frac{1}{2} \int \frac{d x}{x^2+3 x+\frac{5}{2}} \\
& =\frac{1}{2} \int \frac{d x}{\left(x+\frac{3}{2}\right)^2+\left(\frac{1}{2}\right)^2}
\end{aligned}
$
Put $x+\frac{3}{2}=t$, so that $d x=d t$, we get
$
\begin{aligned}
I_2 & =\frac{1}{2} \int \frac{d t}{t^2+\left(\frac{1}{2}\right)^2}=\frac{1}{2 \times \frac{1}{2}} \tan ^{-1} 2 t+C_2 \\
& =\tan ^{-1} 2\left(x+\frac{3}{2}\right)+C_2=\tan ^{-1}(2 x+3)+C_2
\end{aligned}
$
[by $7.4(3)]$
Using (2) and (3) in (1), we get
$
\int \frac{x+2}{2 x^2+6 x+5} d x=\frac{1}{4} \log \left|2 x^2+6 x+5\right|+\frac{1}{2} \tan ^{-1}(2 x+3)+\mathrm{C}
$
where,
$
\mathrm{C}=\frac{\mathrm{C}_1}{4}+\frac{\mathrm{C}_2}{2}
$
(ii) This integral is of the form given in 7.4 (10). Let us express
$
x+3=\mathrm{A} \frac{d}{d x}\left(5-4 x-x^2\right)+\mathrm{B}=\mathrm{A}(-4-2 x)+\mathrm{B}
$
Equating the coefficients of $x$ and the constant terms from both sides, we get $-2 \mathrm{~A}=1$ and $-4 \mathrm{~A}+\mathrm{B}=3$, i.e., $\mathrm{A}=-\frac{1}{2}$ and $\mathrm{B}=1$
Therefore, $\int \frac{x+3}{\sqrt{5-4 x-x^2}} d x=-\frac{1}{2} \int \frac{(-4-2 x) d x}{\sqrt{5-4 x-x^2}}+\int \frac{d x}{\sqrt{5-4 x-x^2}}$
$
=-\frac{1}{2} \mathrm{I}_1+\mathrm{I}_2
$
In $\mathrm{I}_1$, put $5-4 x-x^2=t$, so that $(-4-2 x) d x=d t$.
Therefore,
$
\begin{aligned}
\mathrm{I}_1 & =\int \frac{(-4-2 x) d x}{\sqrt{5-4 x-x^2}}=\int \frac{d t}{\sqrt{t}}=2 \sqrt{t}+\mathrm{C}_1 \\
& =2 \sqrt{5-4 x-x^2}+\mathrm{C}_1
\end{aligned}
$
Now consider
$
\mathrm{I}_2=\int \frac{d x}{\sqrt{5-4 x-x^2}}=\int \frac{d x}{\sqrt{9-(x+2)^2}}
$
Put $x+2=t$, so that $d x=d t$.
Therefore
$
\begin{aligned}
\mathrm{I}_2 & =\int \frac{d t}{\sqrt{3^2-t^2}}=\sin ^{-1} \frac{t}{3}+\mathrm{C}_2 \\
& =\sin ^{-1} \frac{x+2}{3}+\mathrm{C}_2
\end{aligned}
$
Substituting (2) and (3) in (1), we obtain
$
\int \frac{x+3}{\sqrt{5-4 x-x^2}}=-\sqrt{5-4 x-x^2}+\sin ^{-1} \frac{x+2}{3}+\mathrm{C}, \text { where } \mathrm{C}=\mathrm{C}_2-\frac{\mathrm{C}_1}{2}
$
Example 11
Find $\int \frac{d x}{(x+1)(x+2)}$
Solution
The integrand is a proper rational function. Therefore, by using the form of partial fraction [Table 7.2 (i)], we write
$
\frac{1}{(x+1)(x+2)}=\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B}}{x+2}
$
where, real numbers A and B are to be determined suitably. This gives
$
1=\mathrm{A}(x+2)+\mathrm{B}(x+1) \text {. }
$
Equating the coefficients of $x$ and the constant term, we get
and
$
\begin{aligned}
A+B & =0 \\
2 A+B & =1
\end{aligned}
$
Solving these equations, we get $\mathrm{A}=1$ and $\mathrm{B}=-1$. Thus, the integrand is given by
$
\frac{1}{(x+1)(x+2)}=\frac{1}{x+1}+\frac{-1}{x+2}
$
Therefore,
$
\begin{aligned}
\int \frac{d x}{(x+1)(x+2)} & =\int \frac{d x}{x+1}-\int \frac{d x}{x+2} \\
& =\log |x+1|-\log |x+2|+\mathrm{C} \\
& =\log \left|\frac{x+1}{x+2}\right|+\mathrm{C}
\end{aligned}
$
Example 12
Find $\int \frac{x^2+1}{x^2-5 x+6} d x$
Solution
Here the integrand $\frac{x^2+1}{x^2-5 x+6}$ is not proper rational function, so we divide $x^2+1$ by $x^2-5 x+6$ and find that
$
\frac{x^2+1}{x^2-5 x+6}=1+\frac{5 x-5}{x^2-5 x+6}=1+\frac{5 x-5}{(x-2)(x-3)}
$
Let
$
\frac{5 x-5}{(x-2)(x-3)}=\frac{\mathrm{A}}{x-2}+\frac{\mathrm{B}}{x-3}
$
So that
$
5 x-5=\mathrm{A}(x-3)+\mathrm{B}(x-2)
$
Equating the coefficients of $x$ and constant terms on both sides, we get $\mathrm{A}+\mathrm{B}=5$ and $3 A+2 B=5$. Solving these equations, we get $A=-5$ and $B=10$
Thus,
$
\frac{x^2+1}{x^2-5 x+6}=1-\frac{5}{x-2}+\frac{10}{x-3}
$
Therefore,
$
\begin{aligned}
\int \frac{x^2+1}{x^2-5 x+6} d x & =\int d x-5 \int \frac{1}{x-2} d x+10 \int \frac{d x}{x-3} \\
& =x-5 \log |x-2|+10 \log |x-3|+\text { C. }
\end{aligned}
$
Example 13
Find $\int \frac{3 x-2}{(x+1)^2(x+3)} d x$
Solution
The integrand is of the type as given in Table 7.2 (4). We write
$
\frac{3 x-2}{(x+1)^2(x+3)}=\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B}}{(x+1)^2}+\frac{\mathrm{C}}{x+3}
$
So that
$
\begin{aligned}
3 x-2 & =\mathrm{A}(x+1)(x+3)+\mathrm{B}(x+3)+\mathrm{C}(x+1)^2 \\
& =\mathrm{A}\left(x^2+4 x+3\right)+\mathrm{B}(x+3)+\mathrm{C}\left(x^2+2 x+1\right)
\end{aligned}
$
Comparing coefficient of $x^2, x$ and constant term on both sides, we get $A+C=0,4 A+B+2 C=3$ and $3 A+3 B+C=-2$. Solving these equations, we get $\mathrm{A}=\frac{11}{4}, \mathrm{~B}=\frac{-5}{2}$ and $\mathrm{C}=\frac{-11}{4}$. Thus the integrand is given by
$
\begin{aligned}
\frac{3 x-2}{(x+1)^2(x+3)} & =\frac{11}{4(x+1)}-\frac{5}{2(x+1)^2}-\frac{11}{4(x+3)} \\
\int \frac{3 x-2}{(x+1)^2(x+3)} & =\frac{11}{4} \int \frac{d x}{x+1}-\frac{5}{2} \int \frac{d x}{(x+1)^2}-\frac{11}{4} \int \frac{d x}{x+3} \\
& =\frac{11}{4} \log |x+1|+\frac{5}{2(x+1)}-\frac{11}{4} \log |x+3|+\mathrm{C} \\
& =\frac{11}{4} \log \left|\frac{x+1}{x+3}\right|+\frac{5}{2(x+1)}+\mathrm{C}
\end{aligned}
$
Example 14
Find $\int \frac{x^2}{\left(x^2+1\right)\left(x^2+4\right)} d x$
Solution
Consider $\frac{x^2}{\left(x^2+1\right)\left(x^2+4\right)}$ and put $x^2=y$.
Then
$
\frac{x^2}{\left(x^2+1\right)\left(x^2+4\right)}=\frac{y}{(y+1)(y+4)}
$
Write
$
\frac{y}{(y+1)(y+4)}=\frac{\mathrm{A}}{y+1}+\frac{\mathrm{B}}{y+4}
$
So that
$
y=\mathrm{A}(y+4)+\mathrm{B}(y+1)
$
Comparing coefficients of $y$ and constant terms on both sides, we get $\mathrm{A}+\mathrm{B}=1$ and $4 A+B=0$, which give
$
\mathrm{A}=-\frac{1}{3} \text { and } \mathrm{B}=\frac{4}{3}
$
Thus,
$
\frac{x^2}{\left(x^2+1\right)\left(x^2+4\right)}=-\frac{1}{3\left(x^2+1\right)}+\frac{4}{3\left(x^2+4\right)}
$
Therefore,
$
\begin{aligned}
\int \frac{x^2 d x}{\left(x^2+1\right)\left(x^2+4\right)} & =-\frac{1}{3} \int \frac{d x}{x^2+1}+\frac{4}{3} \int \frac{d x}{x^2+4} \\
& =-\frac{1}{3} \tan ^{-1} x+\frac{4}{3} \times \frac{1}{2} \tan ^{-1} \frac{x}{2}+C \\
& =-\frac{1}{3} \tan ^{-1} x+\frac{2}{3} \tan ^{-1} \frac{x}{2}+\mathrm{C}
\end{aligned}
$
In the above example, the substitution was made only for the partial fraction part and not for the integration part. Now, we consider an example, where the integration involves a combination of the substitution method and the partial fraction method.
Example 15
Find $\int \frac{(3 \sin \phi-2) \cos \phi}{5-\cos ^2 \phi-4 \sin \phi} d \phi$
Solution
Let $y=\sin \phi$
Then
$
d y=\cos \phi d \phi
$
Therefore,
$
\begin{aligned}
\int \frac{(3 \sin \phi-2) \cos \phi}{5-\cos ^2 \phi-4 \sin \phi} d \phi & =\int \frac{(3 y-2) d y}{5-\left(1-y^2\right)-4 y} \\
& =\int \frac{3 y-2}{y^2-4 y+4} d y \\
& =\int \frac{3 y-2}{(y-2)^2}=\mathrm{I} \text { (say) }
\end{aligned}
$
Now, we write
$
\frac{3 y-2}{(y-2)^2}=\frac{\mathrm{A}}{y-2}+\frac{\mathrm{B}}{(y-2)^2}
$
[by Table $7.2(2)$ ]
Therefore,
$
3 y-2=\mathrm{A}(y-2)+\mathrm{B}
$
Comparing the coefficients of $y$ and constant term, we get $\mathrm{A}=3$ and $\mathrm{B}-2 \mathrm{~A}=-2$, which gives $\mathrm{A}=3$ and $\mathrm{B}=4$.
Therefore, the required integral is given by
$
\begin{aligned}
\mathrm{I} & =\int\left[\frac{3}{y-2}+\frac{4}{(y-2)^2}\right] d y=3 \int \frac{d y}{y-2}+4 \int \frac{d y}{(y-2)^2} \\
& =3 \log |y-2|+4\left(-\frac{1}{y-2}\right)+\mathrm{C} \\
& =3 \log |\sin \phi-2|+\frac{4}{2-\sin \phi}+\mathrm{C} \\
& \left.=3 \log (2-\sin \phi)+\frac{4}{2-\sin \phi}+\mathrm{C} \text { (since, } 2-\sin \phi \text { is always positive }\right)
\end{aligned}
$
Example 16
Find $\int \frac{x^2+x+1 d x}{(x+2)\left(x^2+1\right)}$
Solution
The integrand is a proper rational function. Decompose the rational function into partial fraction $[$ Table 2.2(5)]. Write
$
\frac{x^2+x+1}{\left(x^2+1\right)(x+2)}=\frac{\mathrm{A}}{x+2}+\frac{\mathrm{B} x+\mathrm{C}}{\left(x^2+1\right)}
$
Therefore,
$
x^2+x+1=\mathrm{A}\left(x^2+1\right)+(\mathrm{B} x+\mathrm{C})(x+2)
$
Equating the coefficients of $x^2, x$ and of constant term of both sides, we get $\mathrm{A}+\mathrm{B}=1,2 \mathrm{~B}+\mathrm{C}=1$ and $\mathrm{A}+2 \mathrm{C}=1$. Solving these equations, we get $\mathrm{A}=\frac{3}{5}, \mathrm{~B}=\frac{2}{5}$ and $\mathrm{C}=\frac{1}{5}$
Thus, the integrand is given by
$
\frac{x^2+x+1}{\left(x^2+1\right)(x+2)}=\frac{3}{5(x+2)}+\frac{\frac{2}{5} x+\frac{1}{5}}{x^2+1}=\frac{3}{5(x+2)}+\frac{1}{5}\left(\frac{2 x+1}{x^2+1}\right)
$
Therefore,
$
\begin{aligned}
\int \frac{x^2+x+1}{\left(x^2+1\right)(x+2)} d x & =\frac{3}{5} \int \frac{d x}{x+2}+\frac{1}{5} \int \frac{2 x}{x^2+1} d x+\frac{1}{5} \int \frac{1}{x^2+1} d x \\
& =\frac{3}{5} \log |x+2|+\frac{1}{5} \log \left|x^2+1\right|+\frac{1}{5} \tan ^{-1} x+\mathrm{C}
\end{aligned}
$
Example 17
Find $\int x \cos x d x$
Solution
Put $f(x)=x$ (first function) and $g(x)=\cos x$ (second function). Then, integration by parts gives
$
\begin{aligned}
\int x \cos x d x & =x \int \cos x d x-\int\left[\frac{d}{d x}(x) \int \cos x d x\right] d x \\
& =x \sin x-\int \sin x d x=x \sin x+\cos x+C
\end{aligned}
$
Suppose, we take
$
f(x)=\cos x \text { and } g(x)=x \text {. Then }
$
$
\begin{aligned}
\int x \cos x d x & =\cos x \int x d x-\int\left[\frac{d}{d x}(\cos x) \int x d x\right] d x \\
& =(\cos x) \frac{x^2}{2}+\int \sin x \frac{x^2}{2} d x
\end{aligned}
$
Thus, it shows that the integral $\int x \cos x d x$ is reduced to the comparatively more complicated integral having more power of $x$. Therefore, the proper choice of the first function and the second function is significant.
Example 18
Find $\int \log x d x$
Solution
To start with, we are unable to guess a function whose derivative is $\log x$. We take $\log x$ as the first function and the constant function 1 as the second function. Then, the integral of the second function is $x$.
Hence,
$
\begin{aligned}
\int(\log x .1) d x & =\log x \int 1 d x-\int\left[\frac{d}{d x}(\log x) \int 1 d x\right] d x \\
& =(\log x) \cdot x-\int \frac{1}{x} x d x=x \log x-x+\mathrm{C} .
\end{aligned}
$
Example 19
Find $\int x e^x d x$
Solution
Take first function as $x$ and second function as $e^x$. The integral of the second function is $e^x$.
Therefore,
$
\int x e^x d x=x e^x-\int 1 \cdot e^x d x=x e^x-e^x+\text { C. }
$
Example 20
Find $\int \frac{x \sin ^{-1} x}{\sqrt{1-x^2}} d x$
Solution
Let first function be $\sin ^{-1} x$ and second function be $\frac{x}{\sqrt{1-x^2}}$.
First we find the integral of the second function, i.e., $\int \frac{x d x}{\sqrt{1-x^2}}$.
Put $t=1-x^2$. Then $d t=-2 x d x$
Therefore, $\quad \int \frac{x d x}{\sqrt{1-x^2}}=-\frac{1}{2} \int \frac{d t}{\sqrt{t}}=-\sqrt{t}=-\sqrt{1-x^2}$
Hence, $\quad \quad \quad \frac{x \sin ^{-1} x}{\sqrt{1-x^2}} d x=\left(\sin ^{-1} x\right)\left(-\sqrt{1-x^2}\right)-\int \frac{1}{\sqrt{1-x^2}}\left(-\sqrt{1-x^2}\right) d x$
$
=-\sqrt{1-x^2} \sin ^{-1} x+x+\mathrm{C}=x-\sqrt{1-x^2} \sin ^{-1} x+\mathrm{C}
$
Alternatively, this integral can also be worked out by making substitution $\sin ^{-1} x=\theta$ and then integrating by parts.
Example 21
Find $\int e^x \sin x d x$
Solution
Take $e^x$ as the first function and $\sin x$ as second function. Then, integrating by parts, we have
$
\begin{aligned}
\mathrm{I}=\int e^x \sin x d x & =e^x(-\cos x)+\int e^x \cos x d x \\
& =-e^x \cos x+\mathrm{I}_1 \text { (say) }
\end{aligned}
$
Taking $e^x$ and $\cos x$ as the first and second functions, respectively, in $I_1$, we get
$
\mathrm{I}_1=e^x \sin x-\int e^x \sin x d x
$
Substituting the value of $I_1$ in (1), we get
$
\mathrm{I}=-e^x \cos x+e^x \sin x-\mathrm{I} \text { or } 2 \mathrm{I}=e^x(\sin x-\cos x)
$
Hence, $\quad \mathrm{I}=\int e^x \sin x d x=\frac{e^x}{2}(\sin x-\cos x)+\mathrm{C}$
Alternatively, above integral can also be determined by taking $\sin x$ as the first function and $e^x$ the second function.
Example 22
Find
(i) $\int e^x\left(\tan ^{-1} x+\frac{1}{1+x^2}\right) d x$
(ii) $\int \frac{\left(x^2+1\right) e^x}{(x+1)^2} d x$
Solution
(i) We have $\mathrm{I}=\int e^x\left(\tan ^{-1} x+\frac{1}{1+x^2}\right) d x$
Consider $f(x)=\tan ^{-1} x$, then $f^{\prime}(x)=\frac{1}{1+x^2}$
Thus, the given integrand is of the form $e^x\left[f(x)+f^{\prime}(x)\right]$.
Therefore, $\mathrm{I}=\int e^x\left(\tan ^{-1} x+\frac{1}{1+x^2}\right) d x=e^x \tan ^{-1} x+\mathrm{C}$
(ii) We have $\mathrm{I}=\int \frac{\left(x^2+1\right) e^x}{(x+1)^2} d x=\int e^x\left[\frac{\left.x^2-1+1+1\right)}{(x+1)^2}\right] d x$
$
=\int e^x\left[\frac{x^2-1}{(x+1)^2}+\frac{2}{(x+1)^2}\right] d x=\int e^x\left[\frac{x-1}{x+1}+\frac{2}{(x+1)^2}\right] d x
$
Consider $f(x)=\frac{x-1}{x+1}$, then $f^{\prime}(x)=\frac{2}{(x+1)^2}$
Thus, the given integrand is of the form $e^x\left[f(x)+f^{\prime}(x)\right]$.
Therefore, $\quad \int \frac{x^2+1}{(x+1)^2} e^x d x=\frac{x-1}{x+1} e^x+\mathrm{C}$
Example 23
Find $\int \sqrt{x^2+2 x+5} d x$
Solution
Note that
$
\int \sqrt{x^2+2 x+5} d x=\int \sqrt{(x+1)^2+4} d x
$
Put $x+1=y$, so that $d x=d y$. Then
$
\begin{aligned}
\int \sqrt{x^2+2 x+5} d x & =\int \sqrt{y^2+2^2} d y \\
& =\frac{1}{2} y \sqrt{y^2+4}+\frac{4}{2} \log \left|y+\sqrt{y^2+4}\right|+\mathrm{C} \quad[\text { using 7.6.2 (ii)] } \\
& =\frac{1}{2}(x+1) \sqrt{x^2+2 x+5}+2 \log \left|x+1+\sqrt{x^2+2 x+5}\right|+\mathrm{C}
\end{aligned}
$
Example 24
Find $\int \sqrt{3-2 x-x^2} d x$
Solution
Note that $\int \sqrt{3-2 x-x^2} d x=\int \sqrt{4-(x+1)^2} d x$
Put $x+1=y$ so that $d x=d y$.
Thus
$
\begin{aligned}
\int \sqrt{3-2 x-x^2} d x & =\int \sqrt{4-y^2} d y \\
& =\frac{1}{2} y \sqrt{4-y^2}+\frac{4}{2} \sin ^{-1} \frac{y}{2}+\mathrm{C} \quad \text { [using 7.6.2 (iii)] } \\
& =\frac{1}{2}(x+1) \sqrt{3-2 x-x^2}+2 \sin ^{-1}\left(\frac{x+1}{2}\right)+\mathrm{C}
\end{aligned}
$
Example 25
Evaluate the following integrals:
(i) $\int_2^3 x^2 d x$
(ii) $\int_4^9 \frac{\sqrt{x}}{\left(30-x^{\frac{3}{2}}\right)^2} d x$
(iii) $\int_1^2 \frac{x d x}{(x+1)(x+2)}$
(iv) $\int_0^{\frac{\pi}{4}} \sin ^3 2 t \cos 2 t d t$
Solution
(i) Let $\mathrm{I}=\int_2^3 x^2 d x$. Since $\int x^2 d x=\frac{x^3}{3}=\mathrm{F}(x)$,
Therefore, by the second fundamental theorem, we get
$
I=F(3)-F(2)=\frac{27}{3}-\frac{8}{3}=\frac{19}{3}
$
(ii) Let $\mathrm{I}=\int_4^9 \frac{\sqrt{x}}{\left(30-x^{\frac{3}{2}}\right)^2} d x$. We first find the anti derivative of the integrand.
Put $30-x^{\frac{3}{2}}=t$. Then $-\frac{3}{2} \sqrt{x} d x=d t$ or $\sqrt{x} d x=-\frac{2}{3} d t$
Thus, $\int \frac{\sqrt{x}}{\left(30-x^{\frac{3}{2}}\right)^2} d x=-\frac{2}{3} \int \frac{d t}{t^2}=\frac{2}{3}\left[\frac{1}{t}\right]=\frac{2}{3}\left[\frac{1}{\left(30-x^{\frac{3}{2}}\right)}\right]=\mathrm{F}(x)$
Therefore, by the second fundamental theorem of calculus, we have
$
\begin{aligned}
\mathrm{I} & =\mathrm{F}(9)-\mathrm{F}(4)=\frac{2}{3}\left[\frac{1}{\left(30-x^{\frac{3}{2}}\right)}\right]_4^9 \\
& =\frac{2}{3}\left[\frac{1}{(30-27)}-\frac{1}{30-8}\right]=\frac{2}{3}\left[\frac{1}{3}-\frac{1}{22}\right]=\frac{19}{99}
\end{aligned}
$
(iii) Let $\mathrm{I}=\int_1^2 \frac{x d x}{(x+1)(x+2)}$
Using partial fraction, we get $\frac{x}{(x+1)(x+2)}=\frac{-1}{x+1}+\frac{2}{x+2}$
So
$
\int \frac{x d x}{(x+1)(x+2)}=-\log |x+1|+2 \log |x+2|=\mathrm{F}(x)
$
Therefore, by the second fundamental theorem of calculus, we have
$
\begin{aligned}
\mathrm{I} & =\mathrm{F}(2)-\mathrm{F}(1)=[-\log 3+2 \log 4]-[-\log 2+2 \log 3] \\
& =-3 \log 3+\log 2+2 \log 4=\log \left(\frac{32}{27}\right)
\end{aligned}
$
(iv) Let $\mathrm{I}=\int_0^{\frac{\pi}{4}} \sin ^3 2 t \cos 2 t d t$. Consider $\int \sin ^3 2 t \cos 2 t d t$
Put $\sin 2 t=u$ so that $2 \cos 2 t d t=d u$ or $\cos 2 t d t=\frac{1}{2} d u$
So
$
\begin{aligned}
\int \sin ^3 2 t \cos 2 t d t & =\frac{1}{2} \int u^3 d u \\
& =\frac{1}{8}\left[u^4\right]=\frac{1}{8} \sin ^4 2 t=\mathrm{F}(t) \text { say }
\end{aligned}
$
Therefore, by the second fundamental theorem of integral calculus
$
\mathrm{I}=\mathrm{F}\left(\frac{\pi}{4}\right)-\mathrm{F}(0)=\frac{1}{8}\left[\sin ^4 \frac{\pi}{2}-\sin ^4 0\right]=\frac{1}{8}
$
Example 26
Evaluate $\int_{-1}^1 5 x^4 \sqrt{x^5+1} d x$
Solution
Put $t=x^5+1$, then $d t=5 x^4 d x$.
Therefore, $\quad \int 5 x^4 \sqrt{x^5+1} d x=\int \sqrt{t} d t=\frac{2}{3} t^{\frac{3}{2}}=\frac{2}{3}\left(x^5+1\right)^{\frac{3}{2}}$
Hence,
$
\begin{aligned}
\int_{-1}^1 5 x^4 \sqrt{x^5+1} d x & =\frac{2}{3}\left[\left(x^5+1\right)^{\frac{3}{2}}\right]_{-1}^1 \\
& =\frac{2}{3}\left[\left(1^5+1\right)^{\frac{3}{2}}-\left((-1)^5+1\right)^{\frac{3}{2}}\right] \\
& =\frac{2}{3}\left[2^{\frac{3}{2}}-0^{\frac{3}{2}}\right]=\frac{2}{3}(2 \sqrt{2})=\frac{4 \sqrt{2}}{3}
\end{aligned}
$
Alternatively, first we transform the integral and then evaluate the transformed integral with new limits.
Let
Note that, when
$
\begin{aligned}
t & =x^5+1 . \text { Then } d t=5 x^4 d x \\
x & =-1, t=0 \text { and when } x=1, t=2
\end{aligned}
$
Therefore
$
\begin{aligned}
& \int_{-1}^1 5 x^4 \sqrt{x^5+1} d x=\int_0^2 \sqrt{t} d t \\
& =\frac{2}{3}\left[t^{\frac{3}{2}}\right]_0^2=\frac{2}{3}\left[2^{\frac{3}{2}}-0^{\frac{3}{2}}\right]=\frac{2}{3}(2 \sqrt{2})=\frac{4 \sqrt{2}}{3} \\
&
\end{aligned}
$
Example 27
Evaluate $\int_0^1 \frac{\tan ^{-1} x}{1+x^2} d x$
Solution
Let $t=\tan ^{-1} x$, then $d t=\frac{1}{1+x^2} d x$. The new limits are, when $x=0, t=0$ and when $x=1, t=\frac{\pi}{4}$. Thus, as $x$ varies from 0 to $1, t$ varies from 0 to $\frac{\pi}{4}$.
Therefore
$
\int_0^1 \frac{\tan ^{-1} x}{1+x^2} d x=\int_0^{\frac{\pi}{4}} t d t\left[\frac{t^2}{2}\right]_0^{\frac{\pi}{4}}=\frac{1}{2}\left[\frac{\pi^2}{16}-0\right]=\frac{\pi^2}{32}
$
Example 28
Evaluate $\int_{-1}^2\left|x^3-x\right| d x$
Solution
We note that $x^3-x \geq 0$ on $[-1,0]$ and $x^3-x \leq 0$ on $[0,1]$ and that $x^3-x \geq 0$ on $[1,2]$. So by $\mathrm{P}_2$ we write
$
\begin{aligned}
\int_{-1}^2\left|x^3-x\right| d x & =\int_{-1}^0\left(x^3-x\right) d x+\int_0^1-\left(x^3-x\right) d x+\int_1^2\left(x^3-x\right) d x \\
& =\int_{-1}^0\left(x^3-x\right) d x+\int_0^1\left(x-x^3\right) d x+\int_1^2\left(x^3-x\right) d x \\
& =\left[\frac{x^4}{4}-\frac{x^2}{2}\right]_{-1}^0+\left[\frac{x^2}{2}-\frac{x^4}{4}\right]_0^1+\left[\frac{x^4}{4}-\frac{x^2}{2}\right]_1^2 \\
& =-\left(\frac{1}{4}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+(4-2)-\left(\frac{1}{4}-\frac{1}{2}\right) \\
& =-\frac{1}{4}+\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+2-\frac{1}{4}+\frac{1}{2}=\frac{3}{2}-\frac{3}{4}+2=\frac{11}{4}
\end{aligned}
$
Example 29
Evaluate $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \sin ^2 x d x$
Solution
We observe that $\sin ^2 x$ is an even function. Therefore, by $\mathrm{P}_7$ (i), we get
$
\begin{aligned}
\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \sin ^2 x d x & =2 \int_0^{\frac{\pi}{4}} \sin ^2 x d x \\
& =2 \int_0^{\frac{\pi}{4}} \frac{(1-\cos 2 x)}{2} d x=\int_0^{\frac{\pi}{4}}(1-\cos 2 x) d x
\end{aligned}
$
$
=\left[x-\frac{1}{2} \sin 2 x\right]_0^{\frac{\pi}{4}}=\left(\frac{\pi}{4}-\frac{1}{2} \sin \frac{\pi}{2}\right)-0=\frac{\pi}{4}-\frac{1}{2}
$
Example 30 Evaluate $\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x$
Solution Let $\mathrm{I}=\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x$. Then, by $\mathrm{P}_4$, we have
or
$
2 \mathrm{I}=\pi \int_0^\pi \frac{\sin x d x}{1+\cos ^2 x}
$
or
$
\mathrm{I}=\frac{\pi}{2} \int_0^\pi \frac{\sin x d x}{1+\cos ^2 x}
$
Put $\cos x=t$ so that $-\sin x d x=d t$. When $x=0, t=1$ and when $x=\pi, t=-1$. Therefore, $\left(\right.$ by $\mathrm{P}_1$ ) we get
$
\begin{aligned}
\mathrm{I} & =\frac{-\pi}{2} \int_1^{-1} \frac{d t}{1+t^2}=\frac{\pi}{2} \int_{-1}^1 \frac{d t}{1+t^2} \\
& =\pi \int_0^1 \frac{d t}{1+t^2} \text { (by } \mathrm{P}_7, \text { since } \frac{1}{1+t^2} \text { is even function) } \\
& =\pi\left[\tan ^{-1} t\right]_0^1=\pi\left[\tan ^{-1} 1-\tan ^{-1} 0\right]=\pi\left[\frac{\pi}{4}-0\right]=\frac{\pi^2}{4}
\end{aligned}
$
Example 31
Evaluate $\int_{-1}^1 \sin ^5 x \cos ^4 x d x$
Solution
Let $\mathrm{I}=\int_{-1}^1 \sin ^5 x \cos ^4 x d x$. Let $f(x)=\sin ^5 x \cos ^4 x$. Then $f(-x)=\sin ^5(-x) \cos ^4(-x)=-\sin ^5 x \cos ^4 x=-f(x)$, i.e., $f$ is an odd function. Therefore, by $\mathrm{P}_7$ (ii), $\mathrm{I}=0$
Example 32
Evaluate $\int_0^{\frac{\pi}{2}} \frac{\sin ^4 x}{\sin ^4 x+\cos ^4 x} d x$
Solution
Let $\mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{\sin ^4 x}{\sin ^4 x+\cos ^4 x} d x$
Then, by $\mathrm{P}_4$
$
\mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{\sin ^4\left(\frac{\pi}{2}-x\right)}{\sin ^4\left(\frac{\pi}{2}-x\right)+\cos ^4\left(\frac{\pi}{2}-x\right)} d x=\int_0^{\frac{\pi}{2}} \frac{\cos ^4 x}{\cos ^4 x+\sin ^4 x} d x
$
Adding (1) and (2), we get
$
2 \mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{\sin ^4 x+\cos ^4 x}{\sin ^4 x+\cos ^4 x} d x=\int_0^{\frac{\pi}{2}} d x=[x]_0^{\frac{\pi}{2}}=\frac{\pi}{2}
$
Hence
$
\mathrm{I}=\frac{\pi}{4}
$
Example 33
Evaluate $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d x}{1+\sqrt{\tan x}}$
Solution Let $\mathrm{I}=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d x}{1+\sqrt{\tan x}}=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x} d x}{\sqrt{\cos x}+\sqrt{\sin x}}$
Then, by $\mathrm{P}_3 \quad \mathrm{I}=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos \left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)} d x}{\sqrt{\cos \left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}+\sqrt{\sin \left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}}$
$
=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x
$
Adding (1) and (2), we get
$
2 \mathrm{I}=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} d x=[x]_{\frac{\pi}{6}}^{\frac{\pi}{3}}=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6} \text {. Hence } \mathrm{I}=\frac{\pi}{12}
$
Example 34
Evaluate $\int_0^{\frac{\pi}{2}} \log \sin x d x$
Solution
Let $\mathrm{I}=\int_0^{\frac{\pi}{2}} \log \sin x d x$
Then, by $\mathrm{P}_4$
$
\mathrm{I}=\int_0^{\frac{\pi}{2}} \log \sin \left(\frac{\pi}{2}-x\right) d x=\int_0^{\frac{\pi}{2}} \log \cos x d x
$
Adding the two values of I, we get
$
\begin{aligned}
2 \mathrm{I} & =\int_0^{\frac{\pi}{2}}(\log \sin x+\log \cos x) d x \\
& =\int_0^{\frac{\pi}{2}}(\log \sin x \cos x+\log 2-\log 2) d x(\text { by adding and subtracting } \log 2) \\
& =\int_0^{\frac{\pi}{2}} \log \sin 2 x d x-\int_0^{\frac{\pi}{2}} \log 2 d x \quad \text { (Why?) }
\end{aligned}
$
Put $2 x=t$ in the first integral. Then $2 d x=d t$, when $x=0, t=0$ and when $x=\frac{\pi}{2}$, $t=\pi$.
Therefore
$
\begin{aligned}
2 \mathrm{I} & =\frac{1}{2} \int_0^\pi \log \sin t d t-\frac{\pi}{2} \log 2 \\
& =\frac{2}{2} \int_0^{\frac{\pi}{2}} \log \sin t d t-\frac{\pi}{2} \log 2\left[\text { by } \mathrm{P}_6 \text { as } \sin (\pi-t)=\sin t\right) \\
& =\int_0^{\frac{\pi}{2}} \log \sin x d x-\frac{\pi}{2} \log 2 \text { (by changing variable } t \text { to } x \text { ) }
\end{aligned}
$
$
=\mathrm{I}-\frac{\pi}{2} \log 2
$
Hence
$
\int_0^{\frac{\pi}{2}} \log \sin x d x=\frac{-\pi}{2} \log 2 .
$
Example 35
Find $\int \cos 6 x \sqrt{1+\sin 6 x} d x$
Solution
Put $t=1+\sin 6 x$, so that $d t=6 \cos 6 x d x$
$
\text { Therefore } \quad \begin{aligned}
\int \cos 6 x \sqrt{1+\sin 6 x} d x & =\frac{1}{6} \int t^{\frac{1}{2}} d t \\
& =\frac{1}{6} \times \frac{2}{3}(t)^{\frac{3}{2}}+\mathrm{C}=\frac{1}{9}(1+\sin 6 x)^{\frac{3}{2}}+\mathrm{C}
\end{aligned}
$
Example 36
Find $\int \frac{\left(x^4-x\right)^{\frac{1}{4}}}{x^5} d x$
Solution
We have $\int \frac{\left(x^4-x\right)^{\frac{1}{4}}}{x^5} d x=\int \frac{\left(1-\frac{1}{x^3}\right)^{\frac{1}{4}}}{x^4} d x$
Put $1-\frac{1}{x^3}=1-x^{-3}=t$, so that $\frac{3}{x^4} d x=d t$
Therefore $\int \frac{\left(x^4-x\right)^{\frac{1}{4}}}{x^5} d x=\frac{1}{3} \int t^{\frac{1}{4}} d t=\frac{1}{3} \times \frac{4}{5} t^{\frac{5}{4}}+\mathrm{C}=\frac{4}{15}\left(1-\frac{1}{x^3}\right)^{\frac{5}{4}}+\mathrm{C}$
Example 37
Find $\int \frac{x^4 d x}{(x-1)\left(x^2+1\right)}$
Solution
We have
$
\begin{aligned}
\frac{x^4}{(x-1)\left(x^2+1\right)} & =(x+1)+\frac{1}{x^3-x^2+x-1} \\
& =(x+1)+\frac{1}{(x-1)\left(x^2+1\right)}
\end{aligned}
$
Now express $\quad \frac{1}{(x-1)\left(x^2+1\right)}=\frac{\mathrm{A}}{(x-1)}+\frac{\mathrm{B} x+\mathrm{C}}{\left(x^2+1\right)}$
So
$
\begin{aligned}
1 & =\mathrm{A}\left(x^2+1\right)+(\mathrm{B} x+\mathrm{C})(x-1) \\
& =(\mathrm{A}+\mathrm{B}) x^2+(\mathrm{C}-\mathrm{B}) x+\mathrm{A}-\mathrm{C}
\end{aligned}
$
Equating coefficients on both sides, we get $\mathrm{A}+\mathrm{B}=0, \mathrm{C}-\mathrm{B}=0$ and $\mathrm{A}-\mathrm{C}=1$, which give $A=\frac{1}{2}, B=C=-\frac{1}{2}$. Substituting values of $A, B$ and $C$ in (2), we get
$
\frac{1}{(x-1)\left(x^2+1\right)}=\frac{1}{2(x-1)}-\frac{1}{2} \frac{x}{\left(x^2+1\right)}-\frac{1}{2\left(x^2+1\right)}
$
Again, substituting (3) in (1), we have
$
\frac{x^4}{(x-1)\left(x^2+x+1\right)}=(x+1)+\frac{1}{2(x-1)}-\frac{1}{2} \frac{x}{\left(x^2+1\right)}-\frac{1}{2\left(x^2+1\right)}
$
Therefore
$
\int \frac{x^4}{(x-1)\left(x^2+x+1\right)} d x=\frac{x^2}{2}+x+\frac{1}{2} \log |x-1|-\frac{1}{4} \log \left(x^2+1\right)-\frac{1}{2} \tan ^{-1} x+\mathrm{C}
$
Example 38
Find $\int\left[\log (\log x)+\frac{1}{(\log x)^2}\right] d x$
Solution
Let $\mathrm{I}=\int\left[\log (\log x)+\frac{1}{(\log x)^2}\right] d x$
$
=\int \log (\log x) d x+\int \frac{1}{(\log x)^2} d x
$
In the first integral, let us take 1 as the second function. Then integrating it by parts, we get
$
\begin{aligned}
\mathrm{I} & =x \log (\log x)-\int \frac{1}{x \log x} x d x+\int \frac{d x}{(\log x)^2} \\
& =x \log (\log x)-\int \frac{d x}{\log x}+\int \frac{d x}{(\log x)^2}
\end{aligned}
$
Again, consider $\int \frac{d x}{\log x}$, take 1 as the second function and integrate it by parts,
we have $\int \frac{d x}{\log x}=\left[\frac{x}{\log x}-\int x\left\{-\frac{1}{(\log x)^2}\left(\frac{1}{x}\right)\right\} d x\right]$
Putting (2) in (1), we get
$
\mathrm{I}=x \log (\log x)-\frac{x}{\log x}-\int \frac{d x}{(\log x)^2}+\int \frac{d x}{(\log x)^2}=x \log (\log x)-\frac{x}{\log x}+\mathrm{C}
$$
Example 39
Find $\int[\sqrt{\cot x}+\sqrt{\tan x}] d x$
Solution
We have
$
\mathrm{I}=\int[\sqrt{\cot x}+\sqrt{\tan x}] d x=\int \sqrt{\tan x}(1+\cot x) d x
$
Put $\tan x=t^2$, so that $\sec ^2 x d x=2 t d t$
or
$
\begin{aligned}
d x & =\frac{2 t d t}{1+t^4} \\
\mathrm{I} & =\int t\left(1+\frac{1}{t^2}\right) \frac{2 t}{\left(1+t^4\right)} d t \\
& =2 \int \frac{\left(t^2+1\right)}{t^4+1} d t=2 \int \frac{\left(1+\frac{1}{t^2}\right) d t}{\left(t^2+\frac{1}{t^2}\right)}=2 \int \frac{\left(1+\frac{1}{t^2}\right) d t}{\left(t-\frac{1}{t}\right)^2+2}
\end{aligned}
$
Put $t-\frac{1}{t}=y$, so that $\left(1+\frac{1}{t^2}\right) d t=d y$. Then
$
\begin{aligned}
\mathrm{I} & =2 \int \frac{d y}{y^2+(\sqrt{2})^2}=\sqrt{2} \tan ^{-1} \frac{y}{\sqrt{2}}+\mathrm{C}=\sqrt{2} \tan ^{-1} \frac{\left(t-\frac{1}{t}\right)}{\sqrt{2}}+\mathrm{C} \\
& =\sqrt{2} \tan ^{-1}\left(\frac{t^2-1}{\sqrt{2} t}\right)+\mathrm{C}=\sqrt{2} \tan ^{-1}\left(\frac{\tan x-1}{\sqrt{2 \tan x}}\right)+\mathrm{C}
\end{aligned}
$
Example 40
Find $\int \frac{\sin 2 x \cos 2 x d x}{\sqrt{9-\cos ^4(2 x)}}$
Solution
Let $\mathrm{I}=\int \frac{\sin 2 x \cos 2 x}{\sqrt{9-\cos ^4 2 x}} d x$
Put $\cos ^2(2 x)=t$ so that $4 \sin 2 x \cos 2 x d x=-d t$
Therefore
$
\mathrm{I}=-\frac{1}{4} \int \frac{d t}{\sqrt{9-t^2}}=-\frac{1}{4} \sin ^{-1}\left(\frac{t}{3}\right)+\mathrm{C}=-\frac{1}{4} \sin ^{-1}\left[\frac{1}{3} \cos ^2 2 x\right]+\mathrm{C}
$
Example 41
Evaluate $\int_{-1}^{\frac{3}{2}}|x \sin (\pi x)| d x$
Solution
Here $f(x)=|x \sin \pi x|=\left\{\begin{array}{l}x \sin \pi x \text { for }-1 \leq x \leq 1 \\ -x \sin \pi x \text { for } 1 \leq x \leq \frac{3}{2}\end{array}\right.$
Therefore
$
\begin{aligned}
\int_{-1}^{\frac{3}{2}}|x \sin \pi x| d x & =\int_{-1}^1 x \sin \pi x d x+\int_1^{\frac{3}{2}}-x \sin \pi x d x \\
& =\int_{-1}^1 x \sin \pi x d x-\int_1^{\frac{3}{2}} x \sin \pi x d x
\end{aligned}
$
Integrating both integrals on righthand side, we get
$
\begin{aligned}
\int_{-1}^{\frac{3}{2}}|x \sin \pi x| d x & =\left[\frac{-x \cos \pi x}{\pi}+\frac{\sin \pi x}{\pi^2}\right]_{-1}^1-\left[\frac{-x \cos \pi x}{\pi}+\frac{\sin \pi x}{\pi^2}\right]_1^{\frac{3}{2}} \\
& =\frac{2}{\pi}-\left[-\frac{1}{\pi^2}-\frac{1}{\pi}\right]=\frac{3}{\pi}+\frac{1}{\pi^2}
\end{aligned}
$
Example 42
Evaluate $\int_0^\pi \frac{x d x}{a^2 \cos ^2 x+b^2 \sin ^2 x}$
Solution
Let $\mathrm{I}=\int_0^\pi \frac{x d x}{a^2 \cos ^2 x+b^2 \sin ^2 x}=\int_0^\pi \frac{(\pi-x) d x}{a^2 \cos ^2(\pi-x)+b^2 \sin ^2(\pi-x)}\left(\right.$ using $\left.\mathrm{P}_4\right)$
$
\begin{aligned}
& =\pi \int_0^\pi \frac{d x}{a^2 \cos ^2 x+b^2 \sin ^2 x}-\int_0^\pi \frac{x d x}{a^2 \cos ^2 x+b^2 \sin ^2 x} \\
& =\pi \int_0^\pi \frac{d x}{a^2 \cos ^2 x+b^2 \sin ^2 x}-\mathrm{I}
\end{aligned}
$
Thus
$
2 \mathrm{I}=\pi \int_0^\pi \frac{d x}{a^2 \cos ^2 x+b^2 \sin ^2 x}
$
or
$
\begin{aligned}
\mathrm{I} & =\frac{\pi}{2} \int_0^\pi \frac{d x}{a^2 \cos ^2 x+b^2 \sin ^2 x}=\frac{\pi}{2} \cdot 2 \int_0^{\frac{\pi}{2}} \frac{d x}{a^2 \cos ^2 x+b^2 \sin ^2 x}\left(\text { using } \mathrm{P}_6\right) \\
& =\pi\left[\int_0^{\frac{\pi}{4}} \frac{d x}{a^2 \cos ^2 x+b^2 \sin ^2 x}+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{d x}{a^2 \cos ^2 x+b^2 \sin ^2 x}\right] \\
& =\pi\left[\int_0^{\frac{\pi}{4}} \frac{\sec ^2 x d x}{a^2+b^2 \tan ^2 x}+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\operatorname{cosec}^2 x d x}{\cot ^2 x+b^2}\right] \\
& =\pi\left[\int_0^1 \frac{d t}{a^2+b^2 t^2}-\int_1^0 \frac{d u}{a^2 u^2+b^2}\right](p u t \tan x=\operatorname{tand} \cot x=u) \\
& =\frac{\pi}{a b}\left[\tan ^{-1} \frac{b t}{a}\right]_0^1-\frac{\pi}{a b}\left[\tan ^{-1} \frac{a u}{b}\right]_1^0=\frac{\pi}{a b}\left[\tan ^{-1} \frac{b}{a}+\tan ^{-1} \frac{a}{b}\right]=\frac{\pi^2}{2 a b}
\end{aligned}
$
