Mathematics

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


ANSWER


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)

Adding (1) and (2):
$$2I=\displaystyle\int_{-\pi/2}^{\pi/2}\log 1 dx = \boxed{0}$$
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