Trigonometric integrals. Evaluate the following integrals. ∫0π(1−cos⁡2x)32dx

Step 1
To evaluate the given integrals , ${\displaystyle {\int }_0^{\pi }{(1-\cos 2x)}^{\frac{3}{2}}dx}$.
Solution:
Let us denote the given integral by I as,
${\displaystyle I={\int }_0^{\pi }{(1-\cos 2x)}^{\frac{3}{2}}dx}$
Since we know the identity as, ${\displaystyle 1-\cos 2x={\sin }^{2}x}$,
So the integral becomes,
${\displaystyle I={\int }_0^{\pi }{\left({\sin }^{2}x\right)}^{\frac{3}{2}}dx}$
So further solving the integral gives,
Step 2
${\displaystyle I={\int }_0^{\pi }\left({\sin }^{3}x\right)dx}$
${\displaystyle ={\int }_0^{\pi }\left({\sin }^{2}x\right)\sin xdx}$
${\displaystyle ={\int }_0^{\pi }(1-{\cos }^{2}x)\left(\sin x\right)dx}$
Now using Substitution method, Put $cosx=t$, we get $-\sin xdx=dt$
Therefore, the given integral changes with upper limit, ${\displaystyle \cos \pi =-1}$ and lower limit, ${\displaystyle \cos 0=1}$ as,
$I=-{\int }_1^{-1}(1-t^2)dt$
${\displaystyle I=-{(t-\frac{t^3}{3})}_1^{-1}}$
Step 3
Further, solving the integral gives,
${\displaystyle I={(\frac{t^3}{3}-t)}_1^{-1}}$
${\displaystyle =(\frac{-1}{3}+1)-(\frac{1}{3}-1)}$
${\displaystyle =\frac{2}{3}+\frac{2}{3}}$
${\displaystyle =\frac{4}{3}}$
Hence, the value of the integral is ${\displaystyle \frac{4}{3}}$.