Mathematics

# Evaluate the following definite integrals :$\displaystyle \int _{0}^{\pi} \dfrac {1}{1+\sin x}dx$

##### SOLUTION
$I=\displaystyle \int _{0}^{\pi} \dfrac {1}{1+\sin x}dx$

$=\displaystyle \int _{0}^{\pi}\dfrac {1-\sin x}{(1+\sin x)(1-\sin x)}dx$

$=\displaystyle \int _{0}^{\pi} \dfrac {1-\sin x}{\cos^2 x}dx$

$\Rightarrow \ I=\displaystyle \int _{0}^{\pi} (\sec^2 x-\sec x\tan x)dx$

$= [\tan x-\sec x]_{0}^{\pi}$

$\Rightarrow \ I=(\tan \pi -\sec \pi )-(\tan 0-\sec 0)=0+1+1=2$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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