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Provide Solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 59

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Answer:  \tan ^{-1}2-\frac{\pi}{4}

Hint: To solve this we assume cosec x and cot x in t

Given:  \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\cos e c x \cot x}{1+\operatorname{cosec}^{2} x} d x


\begin{aligned} &\cos e c x=t \\ & \end{aligned}

\operatorname{cosec} x \cdot \cot x d x=d t \\

\int_{2}^{1} \frac{-d t}{1+t^{2}}=-\left[\tan ^{-1} t\right]_{0}^{1} \\

=-\left(\frac{\pi}{4}-\tan ^{-1} 2\right) \\

=\tan ^{-1} 2-\frac{\pi}{4}


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