Indefinite Integration 1 Question 15
16. The value of $\int _{-\pi / 2}^{\pi / 2} \frac{\sin ^{2} x}{1+2^{x}} d x$ is
(2018 Main)
(a) $\frac{\pi}{8}$
(b) $\frac{\pi}{2}$
(c) $4 \pi$
(d) $\frac{\pi}{4}$
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Solution:
- Key idea Use property $=\int _a^{b} f(x) d x=\int _a^{b} f(a+b-x) d x$
$$ \begin{array}{ll} \text { Let } & I=\int _{-\pi / 2}^{\pi / 2} \frac{\sin ^{2} x}{1+2^{x}} d x \\ \Rightarrow & I=\int _{-\pi / 2}^{\pi / 2} \frac{\sin ^{2} \frac{-\pi}{2}+\frac{\pi}{2}-x}{1+2^{-\frac{\pi}{2}+\frac{\pi}{2}-x}} d x \\ \because \int _a^{b} f(x) d x=\int _a^{b} f(a+b-x) d x \\ \Rightarrow & I=\int _{-\pi / 2}^{\pi / 2} \frac{\sin ^{2} x}{1+2^{-x}} d x \\ \Rightarrow & I=\int _{-\pi / 2}^{\pi / 2} \frac{2^{x} \sin ^{2} x}{2^{x}+1} d x \\ \Rightarrow & 2 I=\int _{-\pi / 2}^{\pi / 2} \sin ^{2} x \frac{2^{x}+1}{2^{x}+1} d x \\ \Rightarrow & 2 I=\int _{-\pi / 2}^{\pi / 2} \sin ^{2} x d x \end{array} $$
$$ \begin{array}{ll} \Rightarrow & 2 I=2 \int _0^{\pi / 2} \sin ^{2} x d x\left[\because \sin ^{2} x \text { is an even function }\right] \\ \Rightarrow & I=\int _0^{\pi / 2} \sin ^{2} x d x \\ \Rightarrow & I=\int _0^{\pi / 2} \cos ^{2} x d x \quad \because \int _0^{a} f(x) d x=\int _0^{a} f(a-x) d x \\ \Rightarrow \quad 2 I=\int _0^{\pi / 2} d x \\ \Rightarrow \quad 2 I=[x] _0^{\pi / 2} \Rightarrow I=\frac{\pi}{4} \end{array} $$
