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

JEE Maths question with a full step-by-step solution.

Question

\displaystyle\int_{0}^{\pi/4}\left(\dfrac{x}{x\sin x+\cos x}\right)^{2}dx

equals

\dfrac{4+\pi}{4-\pi}

\dfrac{5-\pi}{5+\pi}

\dfrac{4-\pi}{4+\pi}

correct

\dfrac{5+\pi}{5-\pi}

Solution

Step 1: Prepare for integration by parts. Note

\dfrac{d}{dx}\big(x\sin x+\cos x\big)=\sin x+x\cos x-\sin x=x\cos x

. Split the integrand as

\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=\dfrac{x\cos x}{(x\sin x+\cos x)^{2}}\,dx

and the derivative above,

v=-\dfrac{1}{x\sin x+\cos x}

. For

u=x\sec x

\dfrac{du}{dx}=\sec x+x\sec x\tan x=\sec x\,(1+x\tan x).

Step 3: Apply integration by parts

\displaystyle\int u\,dv=uv-\int v\,du

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+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.

\dfrac{\sec x\,(1+x\tan x)}{x\sin x+\cos x}=\sec^{2}x.

Step 4: Hence

\displaystyle\int_{0}^{\pi/4}\sec^{2}x\,dx=\big[\tan x\big]_{0}^{\pi/4}=1

, and the boundary term at

\dfrac{\pi}{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

0

. Therefore

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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