#### Need Solution for R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 64 Maths Textbook Solution.

Answer: $\frac{\pi }{32}$

Hint: Use indefinite formula then put the limit to solve this integral

Given:$\int_{0}^{\frac{\pi}{4}}(\tan x+\cot x)^{-2} d x$

Solution:

\begin{aligned} &\int_{0}^{\frac{\pi}{4}}(\tan x+\cot x)^{-2} d x=\int_{0}^{\frac{\pi}{4}} \frac{1}{(\tan x+\cot x)^{2}} d x \\ &=\int_{0}^{\frac{\pi}{4}} \frac{1}{\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right)^{2}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}, \cot \theta=\frac{\cos \theta}{\sin \theta}\right] \\ &=\int_{0}^{\frac{\pi}{4}} \frac{1}{\left(\frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos x}\right)^{2}} d x \end{aligned}

$=\int_{0}^{\frac{\pi}{4}}\left(\frac{\sin x \cos x}{\sin ^{2} x+\cos ^{2} x}\right)^{2} d x$                                                    $\left[\sin ^{2} x+\cos ^{2} x=1\right] \\$

$=\int_{0}^{\frac{\pi}{4}} \frac{(\sin x \cos x)^{2}}{1} d x \\$

$=\int_{0}^{\frac{\pi}{4}} \sin ^{2} x \cos ^{2} x d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \$                                                                $\left[1-\sin ^{2} \theta =\cos ^{2} \theta \right] \\$

$=\int_{0}^{\frac{\pi}{4}} \sin ^{2} x\left(1-\sin ^{2} x\right) d x \\$                                                                $\left[\begin{array}{l} 2 \sin ^{2} \theta=1-\cos 2 \theta \\ \sin ^{2} \theta=\frac{1-\cos 2 \theta}{2} \end{array}\right]$

$=\int_{0}^{\frac{\pi}{4}} \sin ^{2} x d x-\int_{0}^{\frac{\pi}{4}} \sin ^{4} x d x$

\begin{aligned} &=\int_{0}^{\frac{\pi}{4}} \frac{1-\cos 2 x}{2} d x-\int_{0}^{\frac{\pi}{4}}\left(\sin ^{2} x\right)^{2} d x \\ &=\int_{0}^{\frac{\pi}{4}} \frac{1}{2}(1-\cos 2 x) d x-\int_{0}^{\frac{\pi}{4}}\left(\frac{1-\cos 2 x}{2}\right)^{2} d x \\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{4}}(1-\cos 2 x) d x-\frac{1}{4} \int_{0}^{\frac{\pi}{4}}(1-\cos 2 x)^{2} d x \\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{4}} 1 d x-\frac{1}{2} \int_{0}^{\frac{\pi}{4}} \cos 2 x d x-\frac{1}{4} \int_{0}^{\frac{\pi}{4}}\left(1+\cos ^{2} 2 x-2 \cos 2 x\right) d x \\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{4}} 1 d x-\frac{1}{2} \int_{0}^{\frac{\pi}{4}} \cos 2 x d x-\frac{1}{4} \int_{0}^{\frac{\pi}{4}} 1 d x-\frac{1}{4} \int_{0}^{\frac{\pi}{4}} \cos ^{2} 2 x d x-\frac{2}{4} \int_{0}^{\frac{\pi}{4}} \cos 2 x d x \\ &=\left(\frac{1}{2}-\frac{1}{4}\right) \int_{0}^{\frac{\pi}{4}} 1 d x-\left(\frac{1}{2}+\frac{1}{2}\right) \int_{0}^{\frac{\pi}{4}} \cos 2 x d x-\frac{1}{4} \int_{0}^{\frac{\pi}{4}}\left(\frac{1+\cos 4 x}{2}\right) d x \end{aligned}

$\left [ 1+\cos 2\theta =2\cos ^{2}\theta \right ]$

\begin{aligned} &=\left(\frac{2-1}{4}\right) \int_{0}^{\frac{\pi}{4}} 1 d x-\int_{0}^{\frac{\pi}{4}} \cos 2 x d x-\frac{1}{8} \int_{0}^{\frac{\pi}{4}} 1 d x-\frac{1}{8} \int_{0}^{\frac{\pi}{4}} \cos 4 x d x \\ &=\frac{1}{4} \int_{0}^{\frac{\pi}{4}} 1 d x-\frac{1}{8} \int_{0}^{\frac{\pi}{4}} 1 d x-\int_{0}^{\frac{\pi}{4}} \cos 2 x d x-\frac{1}{8} \int_{0}^{\frac{\pi}{4}} \cos 4 x d x \\ &=\left(\frac{1}{4}-\frac{1}{8}\right)_{0}^{\frac{\pi}{4}} 1 d x-\int_{0}^{\frac{\pi}{4}} \cos 2 x d x-\frac{1}{8} \int_{0}^{\frac{\pi}{4}} \cos 4 x d x \\ &=\left(\frac{2-1}{8}\right)[x]_{0}^{\frac{\pi}{4}}-\left[\frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{4}}-\frac{1}{8}\left[\frac{\sin 4 x}{4}\right]_{0}^{\frac{\pi}{4}} \end{aligned}

$\left [ \because \int 1dx=x,\int \cos a\: xdx=\frac{\sin ax}{a} \right ]$

$=\frac{1}{8}\left(\frac{\pi}{4}-0\right)-\frac{1}{2}\left[\sin \frac{2 \pi}{4}-\sin 0\right]-\frac{1}{8}\left[\sin \frac{4 \pi}{4}-\sin 0\right]$

$=\frac{1}{8} \frac{\pi}{4}-\frac{1}{2}\left[\sin \frac{\pi}{2}-0\right]-\frac{1}{8}[\sin \pi-\sin 0]$

$\left [ \because \sin 0=0,\sin \pi =0 \right ]$

$=\frac{\pi}{32}-\frac{1}{2}[0-0]-\frac{1}{8}[0-0]$

$=\frac{\pi}{32}$