The value of the integral $I = \int\limits_{ – \pi }^\pi {\frac{{{{\sin }^4}x}}{{1 + {\pi ^x}}}dx} $ is:

(A) $\frac {3}{4} $
(B) $\frac{{3\pi }}{8}$
(C) $\frac{{3\pi }}{4}$
(D) $\frac{{3\pi }}{16}$

Solution

Let x = -t,

$I = \int\limits_\pi ^{ – \pi } { – \frac{{{{\sin }^4}t}}{{1 + {\pi ^{ – t}}}}dt} $

$I = \int\limits_{ – \pi }^\pi {\frac{{{\pi ^t}{{\sin }^4}t}}{{1 + {\pi ^t}}}dt} = \int\limits_{ – \pi }^\pi {\frac{{{\pi ^x}{{\sin }^4}x}}{{1 + {\pi ^x}}}dx} $

Adding the above with the original given integral,

$2I = \int\limits_{ – \pi }^\pi {\frac{{(1 + {\pi ^x}){{\sin }^4}x}}{{1 + {\pi ^x}}}dx} = \int\limits_{ – \pi }^\pi {{{\sin }^4}xdx} $

$\therefore 2I = 2\int\limits_0^\pi {{{\sin }^4}xdx} $

[$\int\limits_{ – a}^a {f(x)dx} = 2\int\limits_0^a {f(x)dx} $ if f(x) is an even function of x]

$I = 2\int\limits_0^{\pi /2} {{{\sin }^4}xdx} = 2 \times \frac{{(4 – 1)(4 – 3)}}{{4.2}} \times \frac{\pi }{2} = \frac{{3\pi }}{8}$

[$\int\limits_0^{2a} {f(x)dx} = 2\int\limits_0^a {f(x)dx} $, if f(2a-x) = f(x) & Walli’s Formula]

Hence, (B)