Mathematics

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

$\pi$

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

$\int \dfrac{x \sec x}{\tan x+\sec x}dx$

$=-\dfrac{x}{\tan x+\sec x}-\int \dfrac{1}{-\tan x-\sec x}dx$

$=-\dfrac{x}{\tan x+\sec x}+\int \dfrac{1}{\tan x+\sec x}dx$

$=-\dfrac{x}{\tan x+\sec x}+\int \dfrac{\cos x}{1+\sin x}dx$
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substitute $u=\sin x\rightarrow du=\cos x dx$

$=\int \dfrac{1}{u+1}du$

$=\ln (u+1)$

$=\ln (\sin x+1)$

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$=-\dfrac{x}{\tan x+\sec x}+\ln (\sin x+1)+C$

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

$=\left [ -\dfrac{x}{\tan x+\sec x}+\ln (\sin x+1)\right ]_0^{\pi}$

$=\pi -0$

$=\pi$

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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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