How to calculate $\int_0^\frac{\pi}{2}(\sin^3x +\cos^3x) \, \mathrm{d}x$? $$\int_0^\frac{\pi}{2}(\sin^3x +\cos^3x) \, \mathrm{d}x$$
How do I compute this? I tried to do the trigonometric manupulation but I can't get the answer.

$a^3+b^3 = (a+b)(a^2+b^2 -ab)$, 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 $\cos x$.
