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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 am not able to find a relevant replacement here. Please help.
$$I=\displaystyle \int_{0}^{2}x^3 \sqrt{2x-x^2} dx=\displaystyle \int_{0}^{2}x^3\sqrt{1-(x-1)^2} dx$$ 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.
Hint: Look at the limits
Once you've figured out why that doesn't work, you should be able to guess what modification you need to do to that substitution to make it work.
Hint: How will you make the limits match the range?
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?
