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Ππ∫-π2π2log(2-sinx2+sinx) is equal to ______. -

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Question

`int_((-π)/2)^(π/2) log((2 - sinx)/(2 + sinx))` is equal to ______.

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MCQ
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Solution

`int_((-π)/2)^(π/2) log((2 - sinx)/(2 + sinx))` is equal to 0.

Explanation:

Let, I = `int_((-π)/2)^(π/2) log((2 - sin x)/(2 + sin x))dx`

f(x) = `log((2 - sinx)/(2 + sinx))`

∴ f(–x) = `((2 - sin(-x))/(2 + sin(-x)))`

= `log((2 + sinx)/(2 - sinx))`

= `-log((2 - sinx)/(2 + sinx))`

= – f(x)

So, f(x) is an odd function.  

Hence, `int_(-π/2)^(π/2) f(x)dx` = 0  ...`[∵ "If" f(x)  "is an odd function, then" int_-a^a f(x)dx = 0]`

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