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

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

Solution:

\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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