Integration of $\int_0^{1/4} \frac{x^3}{ \sqrt{1-9x^2}} dx$ using trigo substitution 
Evalute $\int_0^{1/4} \frac{x^3}{ \sqrt{1-9x^2}} dx$ using trigo substitution

Because $\sqrt{a^2 - b^2x^2} => x = a/b \sin \theta$
So, Let $x= \frac{1}{3} \sin \theta$
$dx = \frac{1}{3} \cos \theta$
Elimination of roots:
$\sqrt{1-9x^2} = \sqrt{1- \sin^2 \theta} = \sqrt{\cos^2 \theta} = \cos \theta $
The limit of integration is $x=0$ to $x=1/4$
Therefore, the new limit of integration is:
$0 \le \theta \le 0.8481$ This is in the 1st quadrant and therefore $\cos \theta$ is positive
Actual substitution:
$\int_0^{1/4} \frac{x^3}{ \sqrt{1-9x^2}} dx = \int_0^{0.8481} \frac{\frac{1}{9} \sin^3 \theta}{\cos \theta} (\frac{1}{3} \cos \theta)d\theta =\int_0^{0.8481} \frac{\sin^3 \theta}{27} d\theta$
However, the calculated value of the definite integral is incorrect. I was told that my trigo substitution is correct. I cannot find which step I made a careless in.
 A: @Accelerator pointed out the mistake for why OP could not get the answer correctly.
Here is another way to solve without trig substitution:
Let $u=x^2$, the definite integral will now become $$\frac{1}{2}\int_{0}^{1/16}\frac{u}{\sqrt{1-9u}}du$$
Let $v=1-9u$, then it will become
$$\frac{-1}{18}\int_{1}^{7/16}\frac{1-v}{9\sqrt{v}}dv=\frac{-1}{18}\int_{1}^{7/16}(\frac{1}{9\sqrt{v}}-\frac{\sqrt{v}}{9})dv$$
A: Let
$$ I = \int_{0}^{1/4} \, \frac{x^3 \, dx}{\sqrt{1- 9 \, x^2}}. $$
Make the substitution $3 \, x = \sin\theta$ to obtain the following.
Let $ a = \sin^{-1}(1/4)$.
\begin{align}
I &= \int_{0}^{a} \, \frac{\sin^3\theta}{3^3} \, \frac{1}{\sqrt{1-\sin^2\theta}} \, \frac{\cos\theta \, d\theta}{3} \\
&= \frac{1}{3^4} \, \int_{0}^{a} \frac{\sin^3\theta \, \cos\theta \, d\theta}{\cos\theta} \\
&= \frac{1}{3^4} \, \int_{0}^{a} \sin^3\theta \, d\theta \\
&= \frac{1}{3^4 \, 4} \, \int_{0}^{a} ( 3 \, \sin\theta - \sin(3\theta) ) \, d\theta \\
&= \frac{1}{3^5 \, 4} \left[ \cos(3\theta) - 9 \, \cos\theta \right]_{0}^{a} \\
&= \frac{1}{3^5 \, 4} \, \left[8 + \cos\left(3 \, \sin^{-1}\left(\frac{3}{4}\right)\right) - 9 \, \cos\left(\sin^{-1}\left(\frac{3}{4}\right)\right) \right] \\
&= \frac{1}{3^5 \, 4} \, \left[ 8 - \frac{5 \, \sqrt{7}}{16} - 9 \, \frac{\sqrt{7}}{4} \right] \\
&= \frac{1}{3^5 \, 4} \, \left(8 - \frac{41 \, \sqrt{7}}{16} \right).
\end{align}
A straight integration reveals
\begin{align}
I &= \int_{0}^{1/4} \, \frac{x^3 \, dx}{\sqrt{1- 9 \, x^2}} \\
&= - \frac{1}{3^5} \, [ (9 x^2 + 2) \, \sqrt{1 - 9 x^2} ]_{0}^{1/4} \\
&= \frac{1}{3^5 \, 4} \, \left( 8 - \frac{41 \, \sqrt{7}}{16} \right)
\end{align}
as expected.
