_{-π}^{π} \; x x/1+² x \, dx = [AP EAPCET 22 May 2024 Shift 1]

Correct Answer is : π22\;\; π²/22π2

Correct Answer is : π22\;\; π²/22π2

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AP EAPCET 22 May 2024 Shift 1 Solved Paper

Section: Mathematics

Question : 77 of 160

Marks: +1, -0

\int\limits_{-\pi}^{\pi} \; \frac{x \sin x}{1+\cos^2 x} \, dx =

Solution:

Let

I=\int_{-\pi}^{\pi}\frac{x\sin x}{1+\cos^2x}\,dx

Since

x\sin x

is even and

1+\cos^2x

is even, the integrand is even.

Thus

I=2\int_0^{\pi}\frac{x\sin x}{1+\cos^2x}\,dx

Using

\int_0^{\pi}x f(\sin x)\,dx=\frac{\pi}{2}\int_0^{\pi}f(\sin x)\,dx

, with

f(\sin x)=\frac{\sin x}{1+\cos^2x}

I=2\cdot\frac{\pi}{2}\int_0^{\pi}\frac{\sin x}{1+\cos^2x}\,dx=\pi\int_0^{\pi}\frac{\sin x}{1+\cos^2x}\,dx

Put

u=\cos x

, so

du=-\sin x\,dx

Then

\int_0^{\pi}\frac{\sin x}{1+\cos^2x}\,dx=\int_{-1}^{1}\frac{du}{1+u^2}

=\left[\tan^{-1}u\right]_{-1}^{1}=\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)=\frac{\pi}{2}

Therefore

I=\pi\cdot\frac{\pi}{2}=\frac{\pi^2}{2}

Hence the correct option is

(3)

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