The value of _-/2^/2(^2x)/(1+3^x),dx is

Question

The value of $\displaystyle\int_{-\pi/2}^{\pi/2}\frac{\cos^2x}{1+3^x}\,dx$ is

Accepted Answer

By King's rule ($x\to-x$) and adding, since $\dfrac1{1+3^x}+\dfrac1{1+3^{-x}}=1$:

$$2I = \int_{-\pi/2}^{\pi/2}\cos^2x\,dx = \int_{-\pi/2}^{\pi/2}\frac{1+\cos2x}2\,dx = \left[\frac x2+\frac{\sin2x}4\right]_{-\pi/2}^{\pi/2} = \frac\pi2$$

$$I = \frac\pi4$$

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