Mathematics

Solve:$\displaystyle \int_{-\pi/2}^{\pi/2} \log \left(\dfrac{2-\sin x}{2+\sin x}\right)dx$

SOLUTION
Property:
$\displaystyle\int_a^bf(x)dx=\displaystyle\int_a^bf(a+b-x)dx$

$I=\displaystyle\int_{-\pi/2}^{\pi/2}\log\left(\dfrac{2-\sin x}{2+\sin x}\right) dx$    (1)
Using above property:
$I=\displaystyle\int_{-\pi/2}^{\pi/2}\log\left(\dfrac{2-\sin (-x)}{2+\sin (-x)}\right) dx$

$\implies I=\displaystyle\int_{-\pi/2}^{\pi/2}\log\left(\dfrac{2+\sin x}{2-\sin x}\right) dx$    (2)

$2I=\displaystyle\int_{-\pi/2}^{\pi/2}\log 1 dx = \boxed{0}$

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One Word Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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