<s>[INST] Evaluate trigonometric integral $ \int_{0}^{\pi / 2} \frac{x^{3} \cos x }{3 \sin x-\sin 3 x}dx $ Evaluate: $$
\int_{0}^{\pi / 2} \frac{x^{3} \cos x d x}{3 \sin x-\sin 3 x}
$$
Here I can see that the denominator nicely converts into $4\sin^{3}{x}$ so I basically get $$
\int_{0}^{\pi / 2}\left(\frac{x}{\sin x}\right)^{3} \cos x\>{dx}
$$
After that I tried substituting $\sin{x}$ as $u$ but that only complicates the problem further, leaving me with an inverse function to deal with. Also, the King's property is not useful here.
Can anyone provide an alternate approach to this question?
 [/INST] \begin{aligned}
& \int_{0}^{\frac{x}{2}} \frac{x^{2} \cos x}{3 \sin x-\sin 3 x} d x \\
=& \int_{0}^{\frac{\pi}{2}} \frac{x^{3} \cos x}{3 \sin x-3 \sin x+4 \sin ^{3} x} \\
=& \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{x^{3} \cos x}{\sin ^{3} x} d x \\
=&-\frac{1}{8} \int_{0}^{\frac{\pi}{2}} x^{3} d\left(\frac{1}{\sin ^{2} x}\right) \\
=&-\left[\frac{x^{3}}{8 \sin ^{2} x}\right]_{0}^{\frac{\pi}{2}}+\frac{3}{8} \int_{0}^{\frac{\pi}{2}} \frac{x^{2}}{\sin ^{2} x} d x \\
=&-\frac{\pi^{3}}{64}+\frac{3}{8} \int_{0}^{\frac{\pi}{2}} \frac{x^{2}}{\sin ^{2} x} d x \\
=& \frac{3 \pi \ln 2}{8}-\frac{\pi^{3}}{64}\quad (\textrm{ Using } \int_{0}^{\frac{\pi}{2}} \frac{x^{2}}{\sin ^{2} x} d x=\pi \ln 2)\\
\end{aligned}
</s>