Solving $\int_0^{\frac{\pi}{2}}\frac{1}{\sin^{2n}(x) + \cos^{2n}(x)}\:dx$
Solution 1:
I cannot seem to add much here other than a slightly quicker way for you to reach the integral you arrive at after making a tangent half-angle substitution.
Factoring out a $\cos^{2n} x$ term in the denominator of the original integral for $I_n$ we have $$I_n = \int_0^{\frac{\pi}{2}} \frac{\sec^{2n} x}{1 + \tan^{2n} x} \, dx = \int_0^{\frac{\pi}{2}} \frac{\sec^{2n - 2} x}{1 + \tan^{2n} x} \sec^2 x \, dx.$$ On setting $t = \tan x, dt = \sec^2 x \, dx$, we have $$I_n = \int_0^\infty \frac{(1 + t^2)^{n - 1}}{1 + t^{2n}} \, dt.$$