Strange Trig Substitution in Integral Compute the following integral
$$\int _0^{0.6}\:\frac{x^2}{\sqrt{9-25x^2}}\mathrm dx$$
So as I have been doing in the past, I thought the required substitution would have been $x = 3\sin\theta$. But in this case, it is actually $x = \frac{3}{5}\sin\theta$. I am not entirely sure why this is. Is it because the coefficient of the $x^2$ in the denominator is also a square number so we can square root it and place it in the denominator of the substitution? Is this a universal rule?
 A: In order to get rid of the square root in                $$\int _0^{0.6}\:\frac{x^2}{\sqrt{9-25x^2}}dx$$
We have to have $9-25x^2$ as a perfect square. This is achieved by having $25x^2 = 9 \sin^2(\theta)$ which implies  $9-25x^2 = 9 \cos^2(\theta)$, which is a perfect square.
A: Maybe to make things clear for you, you need to perform two substitutions.
$$\begin{align}
J=\int_0^{0.6}\frac{x^2}{\sqrt{9-25x^2}}\,dx
&=\int_0^{0.6}\frac{x^2}{3\sqrt{1-\left(\frac{5}{3}x\right)^2}}\,dx
\end{align}$$
Perform the change of variable $y=\frac{5}{3}x$,
$$\begin{align}
J&=\frac{3}{5}\int_0^{1}\frac{\left(\frac{3}{5}y\right)^2}{3\sqrt{1-y^2}}\,dy\\
&=\frac{9}{125}\int_0^{1}\frac{y^2}{\sqrt{1-y^2}}\,dy
\end{align}$$
Since $0.6\times \frac{5}{3}=1$.
The integral bounds belong to $[-1;1]$ and there is the factor $\dfrac{1}{\sqrt{1-y^2}}$ therefore you know the substitution $y=\sin u$ is a way to complete your computation.
Perform the substitution $y=\sin u$,
$$\begin{align}
J&=\frac{9}{125}\int_0^{1}\frac{\sin^2 u}{\sqrt{1-\sin^2 u}}\times \cos u\,du\\
&=\frac{9}{125}\int_0^{1}\sin^2 u\,du\\
\end{align}$$
The substitution works because,
$$\begin{align} \frac{1}{\sqrt{1-y^2}}\,dy&=\frac{\cos u}{\sqrt{1-\sin^2 u}}\,du\\
&=\,du
\end{align}$$
(the derivative of the sinus function is the cosinus function)
No more denominator.
To achieve the computation you need to linearize $\sin^2 u$.
If you know that 
$$\begin{align}\cos(2u)&=\cos^2 u-\sin^ 2 u\\
&=\left(1-\sin^2 u\right) -\sin^2 u\\
&=1-2\sin^2 u\\
\end{align}$$ 
you are done.
But, there is an algorithmic way to tackle the question.
$$\begin{align}\sin^2 u&=\left(\frac{\text{e}^{iu}-\text{e}^{-iu}}{2i}\right)^2\\
&=-\frac{1}{4}\left(\text{e}^{i2u}+\text{e}^{-i2u}-2\times \text{e}^{iu}\times \text{e}^{-iu}\right)\\
&=-\frac{1}{4}\left(\text{e}^{i2u}+\text{e}^{-i2u}-2\right)\\
&=\frac{1}{2}-\frac{1}{2}\cos(2u)
 \end{align}$$
Since,
$$\displaystyle \sin u=\frac{\text{e}^{iu}-\text{e}^{-iu}}{2i},\cos u=\frac{\text{e}^{iu}+\text{e}^{-iu}}{2},i^2=-1$$
Therefore,
$$\begin{align}J&=\frac{9}{125}\int_0^1 \left(\frac{1}{2}-\frac{1}{2}\cos(2u)\right)\,du\\
&=\frac{9}{125}\Big[\frac{1}{2}u-\frac{\sin(2u)}{4}\Big]_{0}^{\frac{\pi}{2}}\\
&=\frac{9}{125}\times \frac{\pi}{4}\\
&=\boxed{\frac{9}{500}\pi}
\end{align}$$
NB:
$\sqrt{1-x^2}$ does exist if and only if $x\in [-1;1]$
You can compute integral $\displaystyle \int_0^1  \frac{P(x)}{\sqrt{1-x^2}}\,dx$ , $P$ is a polynomial function.
A: Because $x=\frac{3}{5}\sin\theta$ implies $x^2=\frac{9}{25}\sin^2\theta$ and when you substitute that into the the integral, you get $9-9\sin^2\theta$ under the radical. Applying the pythagorean theorem results in $9\cos^2\theta$. (The suggested "move" is all about "facilitating" the pythagorean theorem essentially)
A: Look at $\sqrt{9-25x^2}$ It would be nice to find a substitution that would make the square root go away. So we start by symplifying a bit
$$  \sqrt{9-25x^2} = 3 \sqrt{1 - \dfrac{25}{9}x^2}  $$
Now compare $\sqrt{1 - \dfrac{25}{9}x^2}$ to $\sqrt{1 - \sin^2 \theta} = \cos{\theta}$
It seems that $x = \dfrac 35 \sin \theta$ might be a useful substitution.
Note that the biggest (most useful) domain about the origin for $\sin \theta$ is $\theta \in \left[ -\dfrac {\pi}{2}, \dfrac {\pi}{2} \right]$ since, then, $\cos \theta \in [0, 1]$. So, using  $x = \dfrac 35 \cos\theta$ and getting $\sqrt{1 - \cos^2 \theta} = \sin{\theta}$ might cause problems.
