# Definite Integral Evaluation

Evaluate $\displaystyle \int_{0}^{2}x^3\sqrt{(2x-x^2)} dx$

This kind of problem is solved using tricks like putting $\displaystyle x=\sin^2t$ and/or identities like

\begin{align} \int_{0}^{a}f(x) dx &=\int_{0}^{a}f(a-x) dx \end{align}

\begin{align} \int_{0}^{2a}f(x) dx & = \int_{0}^{a}f(2a-x) dx,\; if \; f(2a-x)=f(x) \\ & = 0,\qquad \qquad \qquad if \; f(2a-x)=-f(x) \\ \end{align}

$$I=\displaystyle \int_{0}^{2}x^3 \sqrt{2x-x^2} dx=\displaystyle \int_{0}^{2}x^3\sqrt{1-(x-1)^2} dx$$ Now setting $$x-1=\sin t$$ we obtain $$I=\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\sin t+1)^3 \cos^2 t dt$$
Why doesn't $x = \sin^2 t$ work? Try to answer that first.
It might help if you rewrite the integral as: $$\int_0^2 x^{7/2}\sqrt{2-x}\,dx$$
If it were $1 - x$, then $x = \sin^2 t$ would've worked, as you would get $1 - \sin^2 t = \cos^2 t$. Here though, it's $2 - x$, and $2 - \sin^2 t = 1 + \cos^2 t$, which doesn't help. How will you get rid of that extra $1$? Can you make that $\cos^2 t$ as well?