<s>[INST] Evaluation of $\int^{\frac{\pi}{2}}_{0}\frac{\cos^6 x}{\sin x+\cos x}dx$ 
Evaluation of $\displaystyle \int^{\frac{\pi}{2}}_{0}\frac{\cos^6 x}{\sin x+\cos x}dx$

$\bf{My\; Try::}$ Let $\displaystyle I = \int^{\frac{\pi}{2}}_{0}\frac{\cos^6 x}{\sin x+\cos x}dx = \int^{\frac{\pi}{2}}_{0}\frac{\sin^6 x}{\sin x+\cos x}dx$ 
Above we have used $\displaystyle \int^{a}_{0}f(x)dx  = \int^{a}_{0}f(a-x)dx$
So $$2I = \int^{\frac{\pi}{2}}_{0}\frac{\sin^6 x+\cos^6 x}{\sin x+\cos x}dx = \int^{\frac{\pi}{2}}_{0}\frac{(\sin^2 x+\cos^2 x)(\sin^4 x+\cos^4 x-\sin^2 x\cos^2 x)}{\sin x+\cos x}dx$$
So $$2I = \int^{\frac{\pi}{2}}_{0}\frac{\sin^4 x+\cos^4 x-\sin^2 x\cos^2 x}{\sin x+\cos x}dx = \int^{\frac{\pi}{2}}_{0}\frac{1-3\sin^2 x\cos^2 x}{\sin x+\cos x}dx$$
Now How can i solve it after that, Help Required, Thanks
 [/INST] A little bit expansion of my above comment. Use $\cos x+\sin x=\sqrt{2}\cos(x-\pi/4) $ and changing variable, we get 
$$I=\int_{-\pi/4}^{\pi/4}\frac{\cos^6(x+\pi/4)}{\sqrt{2}\cos (x)}dx.$$
Since $\cos(x+\pi/4)=\frac{1}{\sqrt{2}}(\cos x-\sin x)$, one has 
$$I=\int_{-\pi/4}^{\pi/4}\frac{(\cos x-\sin x)^6}{(\sqrt 2)^7 \cos x}dx.$$
The following is easy.
</s>