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Need Solution for R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 21 Maths Textbook Solution.

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Answer: \frac{-2}{\sqrt{3}}

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

Given: \int_{\frac{\pi }{3}}^{\frac{\pi }{4}}\left ( \tan x+\cot x \right )^{2}dx

Solution:\int_{\frac{\pi}{3}}^{\frac{\pi}{4}}(\tan x+\cot x)^{2} d x=\int_{\frac{\pi}{3}}^{\frac{\pi}{4}}\left[\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right]^{2} d x \quad\left[\tan \theta=\frac{\sin \theta}{\cos \theta}, \cot \theta=\frac{\cos \theta}{\sin \theta}\right]

=\int_{\frac{\pi}{3}}^{\frac{\pi}{4}}\left[\frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos x}\right]^{2} d x \quad\left[\sin ^{2} x+\cos ^{2} x=1\right]

\begin{aligned} &=\int_{\frac{\pi}{3}}^{\frac{\pi}{4}}\left[\frac{1}{\sin x \cos x}\right]^{2} d x \\ &=\int_{\frac{\pi}{3}}^{\frac{\pi}{4}}\left[\frac{2}{2 \sin x \cos x}\right]^{2} d x \end{aligned}                                                                                        \left [ 2\sin x\cos x=\sin 2x \right ]

\begin{aligned} &=\int_{\frac{\pi}{3}}^{\frac{\pi}{4}}\left[\frac{2}{2 \sin x \cos x}\right]^{2} d x \\ &=\int_{\frac{\pi}{3}}^{\frac{\pi}{4}}\left[\frac{2}{\sin 2 x}\right]^{2} d x \\ &=\int_{\frac{\pi}{3}}^{\frac{\pi}{4}} \frac{4}{\sin ^{2} 2 x} d x \end{aligned}                                                                    \left [ \frac{1}{\sin x} =\cos ecx\right ]

\begin{aligned} &=4 \int_{\frac{\pi}{3}}^{\frac{\pi}{4}} \operatorname{cosec}^{2}2 x d x \\ &=4\left[\frac{-\cot 2 x}{2}\right]_{\frac{\pi}{3}}^{\frac{\pi}{4}} \end{aligned}                                                                    \left[\int \operatorname{cosec}^{2} \theta d \theta=-\cot \theta\right]

\begin{aligned} &=\frac{-4}{2}[\cot 2 x]_{\frac{\pi}{3}}^{\frac{\pi}{4}} \\ &=-2\left[\cot 2 \times \frac{\pi}{4}-\cot 2 \times \frac{\pi}{3}\right] \\ &=-2\left[\cot \frac{\pi}{2}-\cot \frac{2 \pi}{3}\right]=-2\left[\cot \frac{\pi}{2}-\cot \left(\pi-\frac{\pi}{3}\right)\right] \end{aligned}                                                            \left[\begin{array}{l} \cot \frac{\pi}{3}=\frac{1}{\sqrt{3}} \\ \cot \frac{\pi}{2}=0 \end{array}\right]

\begin{aligned} &=-2\left[\cot \frac{\pi}{2}-\left(-\cot \frac{\pi}{3}\right)\right]=-2\left[0-\left(\frac{-1}{\sqrt{3}}\right)\right] \\ &=-2\left(\frac{1}{\sqrt{3}}\right) \\ &=\frac{-2}{\sqrt{3}} \end{aligned}

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