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Question

Question: \[\int_{0}^{\pi}\frac{dx}{1 + \sin x} =\]...

∫0πdx1+sin⁡x=\int_{0}^{\pi}\frac{dx}{1 + \sin x} =

A

0

B

12\frac{1}{2}

C

2

D

32\frac{3}{2}

Answer

2

Explanation

Solution

∫0πdx1+sin⁡x=∫0π1−sin⁡xcos⁡2xdx=∫0π(sec⁡2x−sec⁡xtan⁡x)dx\int_{0}^{\pi}\frac{dx}{1 + \sin x} = \int_{0}^{\pi}{\frac{1 - \sin x}{\cos^{2}x}dx = \int_{0}^{\pi}{(\sec^{2}x - \sec x\tan x)dx}}

=[tan⁡x−sec⁡x]0π=[tan⁡π−sec⁡π+1]=[0+1+1]=2= \lbrack\tan x - \sec x\rbrack_{0}^{\pi} = \lbrack\tan\pi - \sec\pi + 1\rbrack = \lbrack 0 + 1 + 1\rbrack = 2.