Mathematics

# $\displaystyle \underset{0}{\overset{\pi}{\int}} \dfrac{x \, dx}{ 1 + \sin \, x}$ =

$\pi$

##### SOLUTION
$\int \dfrac{x}{1+\sin x}dx$

$=\int \dfrac{x}{\frac{\cos ^2x}{1-\sin x}}dx$

$=\int \left ( \dfrac{x}{\cos ^2x} -\dfrac{x \sin x}{\cos ^2x}\right )dx$

$=x \tan x +\ln|\cos x|-x\sin x \tan x-x \cos x+ \ln| \tan x +\sec x |+C$

$\int_{0}^{\pi}\dfrac{x}{1+\sin x}dx = \left [ x \tan x +\ln|\cos x|-x\sin x \tan x-x \cos x+ \ln| \tan x +\sec x | \right ]_0^{\pi}$

$\int_{0}^{\pi}\dfrac{x}{1+\sin x}dx = \pi$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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