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Please Solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 41 Maths Textbook Solution.

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Answer: $\frac{\pi}{4}$

Hint: In this Statement, we will use $\int_{0}^{a} f(x) d x$ formula

Given: $\int_{0}^{\pi} \frac{x \sin x}{1+\cos ^{2} x} d x$

Solution:

$\int_{0}^{\pi} \frac{x \sin x}{1+\cos ^{2} x} d x$

$\begin{aligned} &\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x \\ & \end{aligned}$

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

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

$\begin{aligned} &=\int_{0}^{\pi} \frac{\pi \sin x}{1+\cos ^{2} x} d x-\int_{0}^{\pi} \frac{x \sin x}{1+\cos ^{2} x} d x \\ & \end{aligned}$

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

$\begin{aligned} &2 I=\pi \int_{0}^{\pi} \frac{\sin x}{1+\cos ^{2} x} d x \\ & \end{aligned}$

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

$\begin{aligned} &I=\frac{\pi}{2} \int_{1}^{-1} \frac{d t}{1+t^{2}} \\ & \end{aligned}$

$I=\frac{\pi}{2}\left[\tan ^{-1} t\right]_{-1}^{1} \\$

$I=\frac{\pi}{2}\left(\tan ^{-1}-\tan ^{-1}(-1)\right. \\$

$I=\frac{\pi}{2}\left[\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)\right]$

$\begin{aligned} &I=\frac{\pi}{2}\left[\frac{\pi}{2}\right] \\ & \end{aligned}$

$I=\frac{\pi}{4}$

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