0
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

$$I=\displaystyle \int_{0}^{2}x^3 \sqrt{2x-x^2} dx=\displaystyle \int_{0}^{2}x^3\sqrt{1-(x-1)^2} dx$$ Now setting $$x-1=\sin t$$ we obtain $$I=\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\sin t+1)^3 \cos^2 t dt$$

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

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?

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.