TITLE: Trick, or Indefinite Integration?
QUESTION [6 upvotes]: Evaluate $$\int_{0}^{\pi/2}\dfrac{\sin^6 x}{\sin x + \cos x}\text{ d}x$$

Is there a nice, elegant way of solving the above integral? 
Here's what I did - 
Replaced $f(x)$ by $f(\pi/2-x)$, added and got rid of some common factors in numerator and denominator. This was followed by indefinite integration, and ultimately substituting the limits in the final expression. 
The above method is kind of long, though - and I'm wondering if there's a quicker, shorter and rather elegant way of getting around this integral. Could someone post a solution, and share ideas? 
Thanks a lot.

REPLY [4 votes]: Hint:
$$2I=\int_0^{\pi/2}\dfrac{\sin^6x+\cos^6x}{\sin x+\cos x}dx$$
Now $\sin^6x+\cos^6x=(\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x(\sin^2x+\cos^2x)=1-3\sin^2x\cos^2x$
and $\sin^2x\cos^2x=\dfrac{\sin^22x}4=\dfrac{1-\cos4x}8$
For $J=\displaystyle\int_0^{\pi/2}\dfrac{\cos4x}{\sin x+\cos x}dx=\int_0^{\pi/2}\dfrac{\cos4x}{\sqrt2\cos\left(x-\dfrac\pi4\right)}dx,$ 
set $x-\dfrac\pi4=y,\cos4x=\cos4\left(y+\dfrac\pi4\right)=\cos(4y+\pi)=-\cos4y$
$\implies-\sqrt2J=\displaystyle\int_{-\pi/4}^{\pi/4}\dfrac{\cos4ydy}{\cos y}=\int_{-\pi/4}^{\pi/4}\dfrac{1-8\cos^2y+8\cos^4y}{\cos y}dy$
Now use $\cos3y=4\cos^3y-3\cos y $