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If I=\int_{0}^{\pi / 2} \frac{\cos x}{1+\sin x} d x . Then find the value of I

Option: 1

2\ln 2


Option: 2

\ln 2


Option: 3

2\ln \frac{\pi}{2}


Option: 4

\ln \frac{\pi}{2}


Answers (1)

best_answer

Let u=1+\sin x 

d u=\cos x d x

\\When\,\,x=0,\,u=1+\sin (0)=1\\ \\When\,\,x=\pi/2,\,\, u=1+\sin \left(\frac{\pi}{2}\right)=2

Then

\int_{0}^{\pi / 2} \frac{\cos x}{1+\sin x}=\int_{1}^{2} u^{-1} d u=\ln |u|_{1}^{2}=[\ln 2-\ln 1]=\ln 2

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SANGALDEEP SINGH

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