#### Provide Solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.3 Question 26

Answer: $-\infty$

Given:  $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{-\frac{\pi}{2}}{\sqrt{\cos x \sin ^{2} x}} d x$

Solution:

Let

\begin{aligned} &I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{-\frac{\pi}{2}}{\sqrt{\cos x \sin ^{2} x}} d x \\ & \end{aligned}

$=-\frac{\pi}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{\sqrt{\cos x \sin ^{2} x}} d x$

\begin{aligned} &=-\frac{\pi}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{\sqrt{\cos x}|\sin x|} d x\\ \end{aligned}

$=-\frac{\pi}{2} \times 2 \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{\cos x}|\sin x|} d x\\$

\begin{aligned} &|\sin x|=\sin x, 0 \leq x \leq \frac{\pi}{2} \\ & \end{aligned}

$=-\pi \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{\cos x} \sin x} d x \\$

$=-\pi \int_{0}^{\frac{\pi}{3}} \frac{1}{\sqrt{\cos x}\left(1-\cos ^{2} x\right)} d x$

Put $\cos x=t^{2}$  and differentiate on both sides

\begin{aligned} &-\sin x d x=2 t d t \\ & \end{aligned}

$\text { when } x \rightarrow 0, t \rightarrow 1 \\$

$\text { when } x \rightarrow \frac{\pi}{2}, t \rightarrow 0$

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

$=2 \pi \int_{1}^{0} \frac{d t}{\left(1-t^{4}\right)} \\$

$=2 \pi \int_{1}^{0} \frac{d t}{(1-t)(1+t)\left(1+t^{2}\right)}$

Now by partial fraction

\begin{aligned} &\frac{1}{(1-t)(1+t)\left(1+t^{2}\right)}=\frac{A}{(1-t)}+\frac{B}{(1+t)}+\frac{C}{\left(1+t^{2}\right)} \\ & \end{aligned}

$1=A(1+t)\left(1+t^{2}\right)+B(1-t)\left(1+t^{2}\right)+C(1-t)(1+t)$

\begin{aligned} &\text { Put } t=1, \text { we get } \\ & \end{aligned}

$A=\frac{1}{4} \\$

$\text { Putting } t=-1, \text { we get }$

\begin{aligned} &1=A+B+D \\ & \end{aligned}

$D=1-\frac{1}{4}-\frac{1}{4} \\$

$=\frac{1}{2}$

Equating coefficients of $t^{3}$ both sides

\begin{aligned} &A-B+C=0 \\ & \end{aligned}

$\frac{1}{4}-\frac{1}{4}+C=0 \\$

$C=0$

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

$=2 \pi \int_{1}^{0} \frac{\frac{1}{4}}{1-t^{2}} d t+2 \pi \int_{1}^{0} \frac{\frac{1}{4}}{1+t} d t+2 \pi \int_{1}^{0} \frac{\frac{1}{2}}{1+t^{2}} d t$

\begin{aligned} &=\frac{2 \pi}{4} \times\left[\frac{\log (1-t)}{-1}\right]_{1}^{0}+\frac{2 \pi}{4} \times[\log (1+t)]_{1}^{0}+\frac{2 \pi}{2} \times\left[\tan ^{-1} t\right]_{1}^{0} \\ & \end{aligned}

$=\frac{-\pi}{2}(\log 1-\log 0)+\frac{\pi}{2}(\log 1-\log 2)+\pi\left(\tan ^{-1} 0-\tan ^{-1} 1\right)$

\begin{aligned} &=\frac{-\pi}{2}[0-(-\infty)]+\frac{\pi}{2}[0-\log 2]+\frac{\pi}{2}\left(0-\frac{\pi}{4}\right) \\ & \end{aligned}

$=-\infty-\frac{\pi}{2} \log 2-\frac{\pi^{2}}{4}$

$= -\infty$ [any validated up with $\infty$ becomes $\infty$]