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AP EAPCET 05-Jul-2022 Shift 2 Paper

Section: Mathematics

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Question : 76 of 160

Marks: +1, -0

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

$2\pi$
$\pi^{2}$
$x+2$
$\frac{\pi}{2}$

Solution: 👈: Video Solution

Let

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

.

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

As

\frac{2x}{1+\cos^{2} x}

is an odd function.

So,

\int\limits_{-\pi}^{\pi} \frac{2x}{1+\cos^{2} x}\,dx=0

As

\frac{2x \sin x}{1+\cos^{2} x}

is an even function.

So,

\int\limits_{-\pi}^{\pi} \frac{2x \sin x}{1+\cos^{2} x}\,dx=2 \int\limits_{0}^{\pi} \frac{2x \sin x}{1+\cos^{2} x}\,dx

I=4 \int\limits_{0}^{\pi} \frac{x \sin x}{1+\cos^{2} x}\,dx

......(i)

=4 \int\limits_{0}^{\pi} \frac{(\pi-x) \sin(\pi-x)}{1+\cos^{2}(\pi-x)}

\Rightarrow I=4 \int\limits_{0}^{\pi} \frac{(\pi-x) \sin x}{1+\cos^{2} x}

.....(ii)

On adding Eqs. (i) and (ii),

2I=4\pi \int\limits_{0}^{\pi} \frac{\sin x}{1+\cos^{2} x}\,dx

Let

\cos x=t

\Rightarrow -\sin x\,dx=dt

When

x \rightarrow 0, t \rightarrow 1

and

x \rightarrow \pi, t \rightarrow -1

I=-2\pi \int\limits_{1}^{-1} \frac{1}{1+t^{2}}\,dt

I=2\pi \int\limits_{-1}^{1} \frac{1}{1+t^{2}}\,dt=2\pi \left[\tan^{-1} t\right]_{-1}^{1}

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

=\pi^{2}

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