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AP EAMCET Engineering 2012 Solved Paper

Section: Mathematics

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Question : 113 of 160

Marks: +1, -0

If

a>0

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

is equal to

$\frac{\pi}{2}$
$\pi$
$\frac{2\pi}{2}$
$a \pi$

Solution:

Let

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

......(i)

Put

x=-x

, we get

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

⇒

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

.......(ii)

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

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

=\int\limits_{-\pi}^{\pi} \sin^{2} x dx

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

=\left[ x+\frac{\sin 2x}{2} \right]_{0}^{\pi}

=\left( \pi+\frac{\sin 2\pi}{2}-0-0 \right)

⇒

2I=\pi

⇒

I=\frac{\pi}{2}

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