印度理工學院(IIT)聯合入學考試-主試(JEE Main)，2018，實體試(offline)，卷1(Paper 1)，A套，第75題

==問題==

計算積分$\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{1 + 2^x} \,{\rm d}x$。

(1) $\displaystyle \frac{\pi}{8}$
(2) $\displaystyle \frac{\pi}{2}$
(3) $\displaystyle 4\pi$
(4) $\displaystyle \frac{\pi}{4}$

==答案==

(4)

==解答==

\begin{align*}
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{1 + 2^x} \,{\rm d}x
&= \int_{0}^{\frac{\pi}{2}} \frac{\sin^2 x}{1 + 2^x} \,{\rm d}x + \int_{-\frac{\pi}{2}}^{0} \frac{\sin^2 x}{1 + 2^x} \,{\rm d}x \\
&= \int_{0}^{\frac{\pi}{2}} \frac{\sin^2 x}{1 + 2^x} \,{\rm d}x + \int_{\frac{\pi}{2}}^{0} \frac{\sin^2 (-t)}{1 + 2^{-t}} (-1) \,{\rm d}t \quad [x: -\frac{\pi}{2} \rightarrow 0, t = -x, t: \frac{\pi}{2} \rightarrow 0] \\
&= \int_{0}^{\frac{\pi}{2}} \frac{\sin^2 x}{1 + 2^x} \,{\rm d}x + \int_{0}^{\frac{\pi}{2}} \frac{\sin^2 x}{1 + 2^{-x}} \,{\rm d}x \\
&= \int_{0}^{\frac{\pi}{2}} \frac{\sin^2 x}{1 + 2^x} \,{\rm d}x + \int_{0}^{\frac{\pi}{2}} \frac{2^x \cdot \sin^2 x}{2^x + 1} \,{\rm d}x \\
&= \int_{0}^{\frac{\pi}{2}} \frac{\sin^2 x (1 + 2^x)}{1 + 2^x} \,{\rm d}x \\
&= \int_{0}^{\frac{\pi}{2}} \sin^2 x \,{\rm d}x \\
&= \int_{0}^{\frac{\pi}{2}} \frac{1 - \cos 2x}{2} \,{\rm d}x \\
&= \frac{1}{2} \left[ x - \frac{1}{2} \sin 2x \right]_{0}^{\frac{\pi}{2}} \\
&= \frac{1}{2} \left[ \left( \frac{\pi}{2} - \frac{1}{2} \sin \pi \right) - (0 - 0) \right] \\
&= \frac{\pi}{4}.
\end{align*}

(解答結束)