Definite integral evaluation of $\frac{\sin^2 x}{2^x + 1}$

You did the right thing. Using your substitution, you have

$$ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{2^x + 1} \ dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{2^x + 1} 2^x \ dx. $$

Adding the two equivalent formulations you have $$ 2 I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{2^x + 1} (2^x +1) \ dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\sin^2 x} \ dx = \int_0^{\frac{\pi}{2}} (1-\cos(2x))\ dx=\frac{\pi}{2}$$ which gives $I = \frac{\pi}{4}$ as the answer.

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Integration

Definite Integrals

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