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Definite Integrals Part 1

Section: Mathematics

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Question : 100 of 100

Marks: +1, -0

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

$\frac{x^{2}}{4}$
$\pi^{2}$
$0$
$\frac{\pi}{2}$

Solution:

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

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

=0+4 \int\limits_{0}^{\pi} \frac{x \sin x dx}{1+\cos^{3}x}

[ as

\int\limits_{-a}^{a} f(x) dx=0

]

if

f(x)

is odd

=2 \int\limits_{0}^{a} f(x) dx

if

f(x)

is even.

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

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

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

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

put

\cos x = t \Rightarrow -\sin x dx = dt

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

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

=2\pi [\tan^{-1} t]_{-1}^{1}

=2\pi [\tan^{-1}1 -\tan^{-1}(-1)]

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

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