Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12

# Π ∫ 0 X Sin X 1 + Cos 2 X D X - Mathematics

Sum

$\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx$

#### Solution

$Let, I = \int_0^\pi \frac{x \sin x}{1 + \cos^2 x} d x ...........(1)$

$= \int_0^\pi \frac{\left( \pi - x \right) \sin\left( \pi - x \right)}{1 + \cos^2 \left( \pi - x \right)} d x$

$= \int_0^\pi \frac{\left( \pi - x \right) \sin x}{1 + \cos^2 x} d x ................(2)$

Adding (1) and (2)

$2I = \int_0^\pi \left[ \frac{x \sin x}{1 + \cos^2 x} + \frac{\left( \pi - x \right) \sin x}{1 + \cos^2 x} \right] d x$

$= \int_0^\pi \frac{\pi \sin x}{1 + \cos^2 x} d x$

$= \pi \left[ - \tan^{- 1} \left( cosx \right) \right]_0^\pi$

$= - \pi\left[ \tan^{- 1} \left( - 1 \right) - \tan^{- 1} \left( 1 \right) \right]$

$= - \pi\left( - \frac{\pi}{4} - \frac{\pi}{4} \right)$

$= \frac{\pi^2}{2}$

$Hence, I = \frac{\pi^2}{4}$

Concept: Definite Integrals Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 20 Definite Integrals
Revision Exercise | Q 41 | Page 122