Integration with variable in numerator $$\int\frac{x^5}{\sqrt{25-x^2}}dx$$
I tried to do it with substitution but couldn't get ride of $x^5$ in the numerator.
 A: This can actually be done without a trigonometric substitution.
Let $u = 25-x^2$. Then, $du = -2x\,dx$, and so, $x\,dx = -\dfrac{1}{2}\,du$. 
Also, $x^2 = 25-u$, and so, $x^4 = (25-u)^2$. 
Hence, $\displaystyle\int\dfrac{x^5}{\sqrt{25-x^2}}\,dx$ $= \displaystyle\int\dfrac{x^4}{\sqrt{25-x^2}}x\,dx$ $= -\dfrac{1}{2}\displaystyle\int\dfrac{(25-u)^2}{\sqrt{u}}\,du$ 
$= -\dfrac{1}{2}\displaystyle\int\dfrac{u^2-50u+625}{\sqrt{u}}\,du$ $= -\dfrac{1}{2}\displaystyle\int u^{3/2}-50u^{1/2}+625u^{-1/2}\,du$, 
which is easy to integrate using the power rule.
A: Hint:
Try the substitution:
$$x=5\sin{u}\implies dx=5\cos{u}\,du$$
$$\therefore \sqrt{25-x^2}=\sqrt{25-25\sin^2{u}}=5\cos{u}$$
$$\therefore\int\frac{x^5}{\sqrt{25-x^2}}\,dx=5\int 625\sin^5{u}\,du=3125\int\sin^5{u}\,du$$
A: I like to motivate trigonometric substitution using a figure. Here you want some trig function in the triangle to be $\sqrt{25-x^2}$; that is most naturally treated as one of the legs. Then the hypotenuse must be $5$ while the other leg is $x$. Put the leg of length $x$ opposite the acute angle $\theta$. Then the figure tells you that $x=5\sin(\theta)$ and $\sqrt{25-x^2}=5\cos(\theta)$. Using these substitutions makes your integral fairly straightforward.
To be a bit specific, since $x=5\sin(\theta)$, you can just replace the $x$ in the numerator with this. You then replace $\sqrt{25-x^2}$ with $5 \cos(\theta)$ and replace $dx$ as usual.
A: Hint
Make the change $t=\sqrt{25-x^2}.$ Note that $$dt=-\frac{x}{t}dx$$ and $$x^4=(25-t^2)^2.$$ Thus
$$\int \frac{x^5}{\sqrt{25-x^2}}dx=-\int (25-t^2)^2dt.$$
