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MHTCET 2016 Math Solved Paper

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Question : 22 of 50

Marks: +1, -0

\int\limits_{-\pi/2}^{\pi/2} \log \left( \frac{2-\sin x}{2+\sin x} \right) \, dx

is equal to

1
3
2
0

Solution:

(d) Let,

I=

\int\limits_{-\pi/2}^{\pi/2} \log \left( \frac{2-\sin x}{2+\sin x} \right) \, dx

f(x)=\log \left( \frac{2-\sin x}{2+\sin x} \right)

∴

f(-x)=\log \left( \frac{2-\sin(-x)}{2+\sin(-x)} \right)

=

\log \left( (2+\sin x)(2-\sin x) \right)

=

-\log \left( \frac{2-\sin x}{2+\sin x} \right)

=

-f(x)

\text{So}, f(x)

is an odd function.

Hence,

\int\limits_{-\pi/2}^{\pi/2} f(x) \, dx

= 0

[∵ If f (x) is an odd function, then

\int\limits_{-a}^{a} f(x) \, dx

= 0]

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