Solve $\int\limits_0^{1/\sqrt{2}} \frac{au^2}{5(1-u^2)^2}du = 1$ for $a$ Problem:
The function $f_U(u) = \frac{au^2}{5(1-u^2)^2}$ is a probability density for the random variable $U$, which is non-zero on the interval $(0, \frac{1}{\sqrt{2}})$. I am supposed to find the value $a$.
I understand that that amounts to solving the equation $\int_0^{\frac{1}{\sqrt{2}}} \frac{au^2}{5(1-u^2)^2}du = 1$ for $a$.
Hint:
The hint I am given is to use integration by parts on the left hand side of $$\int_0^{\frac{1}{\sqrt{2}}} \frac{1}{1-u^2}du = \log(1 + \sqrt2)$$
My Attempt:
I don't understand how to use this hint. I know how to integrate by parts.
If the standard expression for integrating by parts is $st\bigg\rvert_0^{\frac{1}{\sqrt{2}}} - \int_0^{\frac{1}{\sqrt{2}}}tds$, then I guess I am supposed to find the appropriate $s$ and $dt$ in the expression $\dfrac{u^2}{(1-u^2)^2}$? In particular, I am supposed to choose $s$ and $dt$ such that $tds = \dfrac{1}{1-u^2}$?
I am led to believe that this is not a particularly nasty integral, though the online integral calculators do some sort of partial fraction decomposition that looks like a mess.
Another idea I had was to do a trigonometric substitution, but I don't think that will work.
HINT:
Let $u=\sin\theta\implies du=\cos\theta \ d \theta$ $$\int_0^{\frac{1}{\sqrt{2}}} \frac{au^2}{5(1-u^2)^2}du =\int_0^{\pi/4}\frac{a\sin^2\theta }{5\cos^4\theta}\cos\theta d\theta$$
$$=\frac a5\int_0^{\pi/4}\frac{(1-\cos^2\theta) }{\cos^3\theta} d\theta$$
$$=\frac a5\int_0^{\pi/4}(\sec^3\theta-\sec\theta) \ d\theta$$
$$=\frac a5\left(\frac12\sec\theta\tan\theta-\frac12\ln|\sec\theta+\tan\theta|\right)_0^{\pi/4} $$
$$=\frac{a}{10}\left(\sqrt{2}-\ln(\sqrt{2}+1)\right)$$
$$\int\limits_{u=0}^{1/\sqrt{2}} \frac{a u^2}{5 (1-u^2)^2}\ du = \frac{1}{10} a \left(\sqrt{2}-\tanh ^{-1}\left(\frac{1}{\sqrt{2}}\right)\right)$$
Then solve
$$\frac{1}{10} a \left(\sqrt{2}-\tanh ^{-1}\left(\frac{1}{\sqrt{2}}\right)\right) = 1$$
to find:
$$a = \frac{10}{\sqrt{2}-\tanh ^{-1}\left(\frac{1}{\sqrt{2}}\right)}$$