JEE Advanced 2018 Paper 1 · Mathematics · Indefinite and Definite Integrals · hard

Let I=∫0^π/2(xsin xcos x)/(sin ^4x+cos ^4x) dx. Find the value of I.

Numerical answer

Let

I=\displaystyle\int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4x+\cos^4x}\,dx

. Find the value of

I

. Give the answer correct to two decimal places.

Show answer and solution

Answer: 0.62

Let

g(x)=\frac{\sin x\cos x}{\sin^4x+\cos^4x}

; it satisfies

g(\frac\pi2-x)=g(x)

. Using

x\to\frac\pi2-x

I

gives

2I=\frac\pi2\int_0^{\pi/2}g(x)\,dx

, so

I=\frac\pi4\int_0^{\pi/2}g\,dx

. Substitute

u=\sin^2x

du=2\sin x\cos x\,dx

, and

\sin^4x+\cos^4x=u^2+(1-u)^2=2u^2-2u+1

. Then

\int_0^{\pi/2}g\,dx=\frac12\int_0^1\frac{du}{2u^2-2u+1}

. Completing the square,

2u^2-2u+1=\frac{(2u-1)^2+1}{2}

, so this equals

\int_0^1\frac{du}{(2u-1)^2+1}=\left[\tfrac12\arctan(2u-1)\right]_0^1=\frac\pi4

. Hence

I=\frac{\pi}{4}\cdot\frac{\pi}{4}=\frac{\pi^2}{16}\approx0.6169

, which rounds to

0.62

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