Question: $$\int\_0^\frac{\pi}{2}(\sin^3x +\cos^3x) \, \mathrm{d}x$$

How do I compute this? I tried to do the trigonometric manupulation but I can't get the answer.

Answer: 
$a^3+b^3 = (a+b)(a^2+b^2 -ab)$, applying this to $a=\sin x, b=\cos x$, produces (noting that $a^2 + b^2 =1$),

$$I= \int (\sin x +\cos x )dx - \int (\sin x + \cos x)\sin x \cos x dx.$$

The second integral can be further expanded and solved with simple substitutions for $\sin x$ and $\cos x$.
$\sin 3x$=$3\sin x$-$4\sin^3 x$ and
$\cos 3x=4\cos^3 x -3\cos x$
.use this formula to subtitute $\sin^3x$ and $\cos^3 x$ and use simple integration