Q. Let $f(x) = \frac{\sin^{2}{\pi x}}{1 + \pi^{x}}$. Then , $\int_{}^{}{\left\lbrack f(x) + f( - x) \right\rbrack dx}$ is equal to
A) 0
B) $x + c$
C) $\frac{x}{2} - \frac{\cos{\pi x}}{2\pi} + c$
D) $\frac{x}{2} - \frac{\sin{2\pi x}}{4\pi} + c$
Explanation
(d)
$f(x) + f( - x) = \frac{\sin^{2}{\pi x}}{1 + \pi^{x}} + \frac{\sin^{2}{( - \pi x)}}{1 + \pi^{- x}}$
$= \frac{\sin^{2}{\pi x}}{1 + \pi^{x}} + \frac{(\sin^{2}{\pi x)\pi^{x}}}{\pi^{x} + 1}$
$= \sin^{2}{\pi x} = \frac{1 - \cos{2\pi x}}{2}$
$\therefore\int_{}^{}{\left\lbrack f(x) + f( - x) \right\rbrack dx} = \int_{}^{}\left\lbrack \frac{1 - \cos{2\pi x}}{2} \right\rbrack dx$
$= \frac{1}{2}x - \frac{\sin{2\pi x}}{4\pi} + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.