The value of _-/2^/2 ^2x1+2^x,dx is:

Question

The value of $\displaystyle\int_{-\pi/2}^{\pi/2} \dfrac{\sin^2x}{1+2^x}\,dx$ is:

Accepted Answer

$I = \int_{-\pi/2}^{\pi/2} \dfrac{\sin^2x}{1+2^x}\,dx$ (i)

Replacing $x 	o -x$: $I = \int_{-\pi/2}^{\pi/2} \dfrac{2^x\sin^2x}{1+2^x}\,dx$ (ii)

Adding (i) and (ii): $2I = \int_{-\pi/2}^{\pi/2}\sin^2x\,dx = \dfrac12\int_{-\pi/2}^{\pi/2}(1-\cos2x)\,dx$

$I = \dfrac14\left[x+\dfrac{\sin2x}{2}ight]_{-\pi/2}^{\pi/2} = \dfrac14\left[\dfrac{\pi}{2}+\dfrac{\pi}{2}ight] = \dfrac{\pi}{4}$

Solutions · 1