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Question: \(\int_{0}^{\pi/4}{\left( \frac{x}{x\sin x + \cos x} \right)^{2}dx}\) =...

∫0π/4(xxsin⁡x+cos⁡x)2dx\int_{0}^{\pi/4}{\left( \frac{x}{x\sin x + \cos x} \right)^{2}dx} =

A

3+π4−π\frac{3 + \pi}{4 - \pi}

B

4−π3+π\frac{4 - \pi}{3 + \pi}

C

4−π4+π\frac{4 - \pi}{4 + \pi}

D

4+π4−π\frac{4 + \pi}{4 - \pi}

Answer

4−π4+π\frac{4 - \pi}{4 + \pi}

Explanation

Solution

Since ddx(1xsin⁡x+cos⁡x)\frac{d}{dx}\left( \frac{1}{x\sin x + \cos x} \right) = −xcos⁡x(xsin⁡x+cos⁡x)2\frac{- x\cos x}{(x\sin x + \cos x)^{2}}

integration by parts,

I = [−xsec⁡xxsin⁡x+cos⁡x+tan⁡x]0π/4\left\lbrack \frac{- x\sec x}{x\sin x + \cos x} + \tan x \right\rbrack_{0}^{\pi/4} = 4−π4+π\frac{4 - \pi}{4 + \pi}