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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Integral Calculus

Integrate $I= \int \limits_{-\pi}^{\pi} \large\frac{\cos ^2 x}{1+a^x}$$dx$

\[\begin {array} {1 1} (a)\;\frac{3 \pi}{4} \\ (b)\;\frac{\pi}{3} \\ (c)\;\frac{\pi}{2} \\ (d)\;\frac{2 \pi}{3} \end {array}\]

1 Answer

By property $f(x)= f(a+b-x)$
$I'= \int \limits_{-\pi}^{\pi} \large\frac{\cos^2 (\pi - \pi -x)}{1+a^{(\pi -\pi-x)}}$
$I'= \int \limits_{-\pi}^{\pi} \large\frac{\cos^2 x}{1+a^{-x}}$$dx$----(1)
$I'= \int \limits_{-\pi}^{\pi} \large\frac{\cos^2 x.a^x}{1+a^{x}}$$dx$----(2)
By adding (1) and (2) we get,
$2I= \int \limits_{-\pi}^{\pi} \cos^2 x\;dx$
$2I= \int \limits_{-\pi}^{\pi} (1+\cos 2x)\;dx$
$\qquad= \bigg[ \large\frac{x}{2} + \frac{\sin 2x}{4} \bigg]_{-\pi}^{\pi}$
$I= \large\frac{1}{2} \times \pi$
$\quad= \large\frac{\pi}{2}$
answered Dec 13, 2013 by meena.p
 
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