<s>[INST] 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.
 [/INST] @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
$$\frac{-1}{18}\int_{1}^{7/16}\frac{1-v}{9\sqrt{v}}dv=\frac{-1}{18}\int_{1}^{7/16}(\frac{1}{9\sqrt{v}}-\frac{\sqrt{v}}{9})dv$$
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