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: Hint: $\sin^3 x dx = -\sin^2 xd(\cos x) = (\cos^2 x - 1)d(\cos x)$. Do the same for $\cos^3 x dx$.
A: $a^3+b^3 = (a+b)(a^2+b^2 -ab)$, 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 $\cos x$.
A: $\int\sin^3xdx=\int\sin^2x\sin(x)dx=-\int(1-\cos^2x)d\cos x$
A: Applying this should give you your answer in no time! 
A: Hint: Use partial integration a couple of times.
A: $\sin 3x$=$3\sin x$-$4\sin^3 x$ 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
A: The curve of $sinx$ and $cosx$ in the interval $(0,\frac{\pi}{2})$ is symmetric. So the curve of $sin^3x$ and $cos^3x$ are symmetric too. Hence the area enclosed under the curves are also equal. So the expression can be re-written as,
$2\int_0^\frac{\pi}{2}sin^3x~dx$ or $2\int_0^\frac{\pi}{2}cos^3x~dx$ whichever you prefer to evaluate.
I proceeded with $sin^3x$.
$2\int_0^\frac{\pi}{2}sin^3x~dx$
$=2\int_0^\frac{\pi}{2}2sinx(1-cos^2x)~dx$
$=2\int_0^\frac{\pi}{2}sinx~dx-2\int_0^\frac{\pi}{2}cos^2xsinx~dx$
Evaluating the first part of the integral is trivial. In the second part substitute $cosx=t$.
Rest of the part is very trivial to calculate. Don't forget to change the limit of the integral after substitution or simply reverse back the substitution you made to the expression after evaluation of the integral.
