#### Please solve RD Sharma class 12 Chapter Definite integrals exercise 19.2 question 40 maths textbook solution.

Answer   : $\frac{\pi}{6}$

Hint         : use indefinite integral formula and the limits to solve this integral

Given      : $\int_{0}^{\frac{\pi}{2}}\frac{\cos ^2 x}{1 +3 \sin ^2 x}$

Solution : $\int_{0}^{\frac{\pi}{2}}\frac{\cos ^2 x}{1 +3 \sin ^2 x}$

on multiplying and dividing by sec4x , we get

\begin{aligned} &=\int_{0}^{\frac{\pi}{2}}\left(\frac{\cos ^{2} x \cdot \sec ^{4} x}{\sec ^{4} x\left(1+3 \sin ^{2} x\right)}\right) d x=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x \cdot \frac{1}{\cos ^{2} x} \cdot \sec ^{2} x}{\sec ^{2} x \cdot \frac{1}{\cos ^{2} x}\left(1+3 \sin ^{2} x\right)} d x \\ &=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x}{\sec ^{2} x\left(\frac{1}{\cos ^{2} x}+3 \frac{\sin ^{2} x}{\cos ^{2} x}\right)} d x \end{aligned}
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x}{\sec ^{2} x\left(\sec ^{2} x+3 \tan ^{2} x\right)} d x \\ &=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x}{\left(1+\tan ^{2} x\right)\left(1+\tan ^{2} x+3 \tan ^{2} x\right)} d x \\ &=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x}{\left(1+\tan ^{2} x\right)\left(1+4 \tan ^{2} x\right)} d x \end{aligned}
put $\tan x=t \Rightarrow \sec^2x dx=dt$
when x=0 then t=0 , when $x =\frac{\pi}{2}$ then $t=\infty$
therefore ,

\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x}{\left(1+\tan ^{2} x\right)\left(1+4 \tan ^{2} x\right)} d x\\ &=\int_{0}^{\infty} \frac{1}{\left(1+t^{2}\right)\left(1+4 t^{2}\right)} d t \end{aligned}

To solve this integral, first we need to find its partial fraction then integrate  it using indefinite integral formula then put the limits to get required answer.
therefore ,

\begin{aligned} &\frac{1}{\left(1+t^{2}\right)\left(1+4 t^{2}\right)}=\frac{A t+B}{\left(1+t^{2}\right)}+\frac{c t+D}{1+4 t^{2}} \\ &\Rightarrow 1=\frac{(A t+B)\left(1+t^{2}\right)\left(1+4 t^{2}\right)}{\left(1+t^{2}\right)}+\frac{(C t+D)\left(1+t^{2}\right)\left(1+4 t^{2}\right)}{\left(1+4 t^{2}\right)} \\ &\Rightarrow 1=(A t+B)\left(1+4 t^{2}\right)+(C t+D)\left(1+t^{2}\right) \\ &\Rightarrow 1=A t+4 A t^{3}+B+4 B t^{2}+C t+C t^{3}+D+D t^{2} \\ &\Rightarrow 1=(4 A+C) t^{3}+(4 B+D) t^{2}+(A+C) t+(B+D) \end{aligned}

Equating the coefficient of t3,t2,t and constant term respectively then

$0=4A+C$             … a
$0=4B+D$            … b
$0=A+C$               … c
$1=B+D$             …(d)
-4A=C
-4B=D

Since A= -C=0
Substracting (d) by (b), we get
4B+D=0
B+D   =13B= -1
B=-13

1= -13+D ? 1+13=D?3+13=D

Therefore $A=C=0, B=-\frac{1}{3} , D=\frac{4}{3}$
Therefore  $\frac{1}{\left.(1+t^{2}\right)\left(1+4 t^{2}\right)}=\frac{0 . t-\frac{1}{3}}{1+t^{2}}+\frac{0 . t+\frac{4}{3}}{1+4 t^{2}}=\frac{-1}{3\left(1+t^{2}\right)}+\frac{4}{3\left(1+4 t^{2}\right)}$
Now (i)?

\begin{aligned} &\int_{0}^{\infty} \frac{1}{\left(1+t^{2}\right)\left(1+4 t^{2}\right)} d t \\ &=\int_{0}^{\infty}\left\{-\frac{1}{3} \frac{1}{\left(1+t^{2}\right)}+\frac{4}{3\left(1+4 t^{2}\right)}\right\} d t \\ &=\frac{-1}{3} \int_{0}^{\infty} \frac{1}{1+t^{2}} d t+\frac{4}{3} \int_{0}^{\infty} \frac{1}{1+4 t^{2}} d t \\ &=\frac{-1}{3} \int_{0}^{\infty} \frac{1}{1+t^{2}} d t+\frac{4}{3} \int_{0}^{\infty} \frac{1}{1+(2 t)^{2}} d t \end{aligned}

put 2t=u ⇒2dt=du ⇒ dt=du2 in second integral, then

\begin{aligned} &\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{1+3 \sin ^{2} x} d x=\frac{-1}{3}\left[\tan ^{-1} t\right]_{0}^{\infty}+\frac{4}{3} \int_{0}^{\infty} \frac{1}{1+u^{2}} \frac{d u}{2} \\ &=\frac{-1}{3}\left[\tan ^{-1} \infty-\tan ^{-1} 0\right]+\frac{2}{3}\left[\tan ^{-1} u\right]_{0}^{\infty} \\ &=\frac{-1}{3}\left[\frac{\pi}{2}-0\right]+\frac{2}{3}\left[\tan ^{-1}(2 t)\right]_{0}^{\infty} \\ &=\frac{-1}{3} \frac{7}{2}+\frac{2}{3}\left[\tan ^{-1} \infty-\tan ^{-1} 2 \times 0\right] \\ &=\frac{-\pi}{6}+\frac{2}{3}\left[\tan ^{-1} \infty-\tan ^{-1} 0\right] \end{aligned}

\begin{aligned} &=\frac{-\pi}{6}+\frac{2}{3}\left[\frac{\pi}{2}-0\right] \\ &=\frac{-\pi}{6}+\frac{2}{3} \frac{\pi}{2} \\ &=\frac{-\pi}{6}+\frac{\pi}{3}=\frac{-\pi+2 \Pi}{6} \\ &=\frac{\pi}{6} \end{aligned}