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∫ (x/(x sinx+cosx))² dx, 0 to π/4 | JEE

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Question
∫0π/4(xxsin⁡x+cos⁡x)2dx\displaystyle\int_{0}^{\pi/4}\left(\dfrac{x}{x\sin x+\cos x}\right)^{2}dx equals
A4+π4−π\dfrac{4+\pi}{4-\pi}
B5−π5+π\dfrac{5-\pi}{5+\pi}
C4−π4+π\dfrac{4-\pi}{4+\pi}correct
D5+π5−π\dfrac{5+\pi}{5-\pi}
Solution
Step 1: Prepare for integration by parts. Note ddx(xsin⁡x+cos⁡x)=sin⁡x+xcos⁡x−sin⁡x=xcos⁡x\dfrac{d}{dx}\big(x\sin x+\cos x\big)=\sin x+x\cos x-\sin x=x\cos x. Split the integrand as
(xxsin⁡x+cos⁡x)2=xcos⁡x⏟u⋅xcos⁡x(xsin⁡x+cos⁡x)2⏟dv.\left(\dfrac{x}{x\sin x+\cos x}\right)^{2}=\underbrace{\dfrac{x}{\cos x}}_{u}\cdot\underbrace{\dfrac{x\cos x}{(x\sin x+\cos x)^{2}}}_{dv}.
 Step 2: With dv=xcos⁡x(xsin⁡x+cos⁡x)2 dxdv=\dfrac{x\cos x}{(x\sin x+\cos x)^{2}}\,dx and the derivative above, v=−1xsin⁡x+cos⁡xv=-\dfrac{1}{x\sin x+\cos x}. For u=xsec⁡xu=x\sec x,
dudx=sec⁡x+xsec⁡xtan⁡x=sec⁡x (1+xtan⁡x).\dfrac{du}{dx}=\sec x+x\sec x\tan x=\sec x\,(1+x\tan x).
 Step 3: Apply integration by parts ∫u dv=uv−∫v du\displaystyle\int u\,dv=uv-\int v\,du:
I=[−xsec⁡xxsin⁡x+cos⁡x]0π/4+∫0π/4sec⁡x (1+xtan⁡x)xsin⁡x+cos⁡x dx.I=\left[-\dfrac{x\sec x}{x\sin x+\cos x}\right]_{0}^{\pi/4}+\int_{0}^{\pi/4}\dfrac{\sec x\,(1+x\tan x)}{x\sin x+\cos x}\,dx.
 The remaining integrand simplifies since sec⁡x(1+xtan⁡x)=cos⁡x+xsin⁡xcos⁡2x=(xsin⁡x+cos⁡x)sec⁡2x1⋅1xsin⁡x+cos⁡x⋅(xsin⁡x+cos⁡x)\sec x(1+x\tan x)=\dfrac{\cos x+x\sin x}{\cos^{2}x}=\dfrac{(x\sin x+\cos x)\sec^{2}x}{1}\cdot\dfrac{1}{x\sin x+\cos x}\cdot(x\sin x+\cos x), i.e.
sec⁡x (1+xtan⁡x)xsin⁡x+cos⁡x=sec⁡2x.\dfrac{\sec x\,(1+x\tan x)}{x\sin x+\cos x}=\sec^{2}x.
 Step 4: Hence ∫0π/4sec⁡2x dx=[tan⁡x]0π/4=1\displaystyle\int_{0}^{\pi/4}\sec^{2}x\,dx=\big[\tan x\big]_{0}^{\pi/4}=1, and the boundary term at π4\dfrac{\pi}{4} is
−π42π4⋅12+12=−π42⋅2π4+1=−π2π+44=−2ππ+4,-\dfrac{\frac{\pi}{4}\sqrt2}{\frac{\pi}{4}\cdot\frac{1}{\sqrt2}+\frac{1}{\sqrt2}}=-\dfrac{\frac{\pi}{4}\sqrt2\cdot\sqrt2}{\frac{\pi}{4}+1}=-\dfrac{\frac{\pi}{2}}{\frac{\pi+4}{4}}=-\dfrac{2\pi}{\pi+4},
 with the lower limit contributing 00. Therefore
I=−2ππ+4+1=(π+4)−2ππ+4=4−π4+π.I=-\dfrac{2\pi}{\pi+4}+1=\dfrac{(\pi+4)-2\pi}{\pi+4}=\dfrac{4-\pi}{4+\pi}.
 Correct answer: (3)
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