JEE Advanced 2016 · Paper 2 · Mathematics Q40

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MathematicsSingle correct+3 / -1Question 40 of 54

The value of $\displaystyle\int_{-\pi/2}^{\pi/2}\frac{x^2\cos x}{1+e^x}\,dx$ is equal to

\dfrac{\pi^2}{4}-2

\dfrac{\pi^2}{4}+2

\pi^2-e^{\pi/2}

\pi^2+e^{\pi/2}

Answer and worked solution

Official answer: A

Fold the interval. Add the integrand at $x$ and $-x$ :
$\frac{x^2\cos x}{1+e^x}+\frac{x^2\cos x}{1+e^{-x}}=x^2\cos x.$
So
$I=\int_0^{\pi/2}x^2\cos x\,dx.$

Integrate by parts.
$I=\Big[x^2\sin x+2x\cos x-2\sin x\Big]_0^{\pi/2}=\frac{\pi^2}4-2.$

(A)

Topics:Properties of definite integrals Integration by parts

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JEE Advanced 2016 · Paper 2, Mathematics question 40