Integral with trigonometric functions

[Tolaso J Kos]

Prove that:

$$\int_{0}^{\pi/4}\int_{\pi/2}^{\pi}\frac{(\cos x-\sin x)^{y-2}}{(\cos x+\sin x)^{y+2}}\, {\rm d}y\, {\rm d}x= \frac{1}{4}\left [ \ln 4 +\ln \left ( \pi^2-1 \right )-\ln \left ( \pi^2-4 \right ) \right ] $$

[galactus]

Wow!, looks like a tricky one with the integration in the exponent.

But, write as:

$$\int_{0}^{\pi/4}\int_{\pi/2}^{\pi}\frac{\sin^{y-2}\left(\frac{\pi}{4}-x\right)}{\sin^{y+2}\left(\frac{\pi}{4}+x\right)}dydx$$

Switch integral signs:

$$\int_{\pi/2}^{\pi}\int_{0}^{\pi/4}\frac{\sin^{y-2}\left(\frac{\pi}{4}-x\right)}{\sin^{y+2}\left(\frac{\pi}{4}+x\right)}dxdy$$

Let $u=\frac{\pi}{4}-x$

$$4\int_{\pi/2}^{\pi}\int_{0}^{\pi/4}\tan^{y}(u)\cdot \frac{1}{\sin^{2}(y)\cos^{2}(y)}dudy$$

$$=4\int_{\pi/2}^{\pi}\int_{0}^{\pi/4}\tan^{y}(u)\csc^{2}(2u)dudy$$

Let $u=\tan^{-1}(t), \;\ du=\frac{1}{t^{2}+1}dt$

$$\int_{\pi/2}^{\pi}\int_{0}^{1}t^{y-2}(t^{2}+1)dtdy$$

$$\int_{\pi/2}^{\pi}\left(\frac{1}{y+1}+\frac{1}{y-1}\right)dy$$

$$=\log\left(4\frac{(\pi^{2}-1)}{\pi^{2}-4}\right)$$