The value of _-^ ^2x1+a^x,dx, a0, is:

Question

The value of $\displaystyle\int_{-\pi}^\pi \dfrac{\cos^2x}{1+a^x}\,dx$, $a>0$, is:

Accepted Answer

Let $I = \int_{-\pi}^\pi \dfrac{\cos^2x}{1+a^x}\,dx$ (1)

$I = \int_{-\pi}^\pi \dfrac{a^x\cos^2x}{1+a^x}\,dx$ (2), using $\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx$

Adding (1) and (2): $2I = \int_{-\pi}^\pi \cos^2x\,dx = 2\int_0^\pi\cos^2x\,dx = 4\int_0^{\pi/2}\cos^2x\,dx = 4\int_0^{\pi/2}\sin^2x\,dx$

$I = 2\int_0^{\pi/2}\sin^2x\,dx = 2\int_0^{\pi/2}(1-\cos2x)\cdot\dfrac12\,dx$

$I = \int_0^{\pi/2}dx - \int_0^{\pi/2}\cos2x\,dx$

$I+I = \dfrac{\pi}{2} \cdot 2 $ *(working through the standard reduction of $\int_0^{\pi/2}\cos^2x\,dx=\pi/4$ gives)* $I = \dfrac{\pi}{2}$

Solutions · 1