The value of $\int_{-\pi / 2}^{\pi / 2} \frac{\sin ^{2} x}{1+2^{x}} d x$ is\

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Professor PC Thomas Classes · Answer

(a) : Let $I=\int_{-\pi / 2}^{\pi / 2} \frac{\sin ^{2} x}{1+2^{x}} d x \quad \ldots$ (i)\ Changing $x$ to $-\pi / 2+\pi / 2-x=-x$, we have\ $I=\int_{-\pi / 2}^{\pi / 2} \frac{\sin ^{2} x}{1+2^{-x}} d x$\ Adding (i) & (ii), we get\ $2 I=\int_{-\pi / 2}^{\pi / 2} \sin ^{2} x\left\{\frac{1}{1+2^{x}}+\frac{1}{1+2^{-x}}ight\} d x$\ $=\int_{-\pi / 2}^{\pi / 2} \sin ^{2} x d x=2 \int_{0}^{\pi / 2} \sin ^{2} x d x$\ (As the integral function is even)\ $	herefore I=\frac{\pi}{4}$\