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
 A: 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.
A: Let
$$　A=\int^{\frac{\pi}{2}}_{0}\frac{\cos^6 x}{\sin x+\cos x}dx, B=\int^{\frac{\pi}{2}}_{0}\frac{\sin^6 x}{\sin x+\cos x}dx. $$
Clearly $A=B$ by changing variable from $x$ to $\frac{\pi}{2}-x$. Now
\begin{eqnarray}
A+B&=&\int^{\frac{\pi}{2}}_{0}\frac{\cos^6 x+\sin^6x}{\sin x+\cos x}dx\\
&=&\int^{\frac{\pi}{2}}_{0}\frac{(\cos^2 x+\sin^2x)(\cos^4-\cos^2x\sin ^2x+\sin^4x)}{\sin x+\cos x}dx\\
&=&\int^{\frac{\pi}{2}}_{0}\frac{1-3\cos^2x\sin ^2x}{\sin x+\cos x}dx\\
&=&\int^{\frac{\pi}{2}}_{0}\frac{1-\frac34\sin ^2(2x)}{\sqrt{2}\sin(x+\frac{\pi}{4})}dx\\
&=&\int^{\frac{3\pi}{24}}_{\frac{\pi}{4}}\frac{1-\frac34\sin ^2[2(u-\frac{\pi}{4})]}{\sqrt{2}\sin u}du\\
&=&\int^{\frac{3\pi}{24}}_{\frac{\pi}{4}}\frac{1-\frac34\cos ^2(2u)}{\sqrt{2}\sin u}du\\
&=&\frac1{4\sqrt2}\int^{\frac{3\pi}{4}}_{\frac{\pi}{4}}\frac{4-3\cos ^2(2u)}{\sin u}du\\
&=&\frac1{4\sqrt2}\int^{\frac{3\pi}{4}}_{\frac{\pi}{4}}\frac{4-3(1-2\sin ^2u)^2}{\sin u}du\\
&=&\frac1{4\sqrt2}\int^{\frac{3\pi}{4}}_{\frac{\pi}{4}}(\frac{1}{\sin u}+12\sin ^2u-12\sin^4 u)du\\
&=&\frac1{4\sqrt2}\left[\ln\tan(\frac{u}{2})-3\cos u-3\cos (3u)\right]\bigg|^{\frac{3\pi}{4}}_{\frac{\pi}{4}}\\
&=&\frac12+\frac1{2\sqrt{2}}\ln(\cot(\frac{\pi}{8}))\\
&=&\frac12+\frac1{2\sqrt{2}}\coth^{-1}(\frac{1}{\sqrt2})
\end{eqnarray}
and hence
$$ A=B=\frac14+\frac1{4\sqrt{2}}\coth^{-1}(\frac{1}{\sqrt2}). $$
