<s>[INST] Evaluate $\int\frac{x^3}{\sqrt{81x^2-16}}dx$ using Trigonometric Substitution $$\int\frac{x^3}{\sqrt{81x^2-16}}dx$$
I started off doing $$u =9x$$ to get $$\int\frac{\frac{u}{9}^3}{\sqrt{u^2-16}}dx$$
which allows for the trig substitution of $$x = a\sec\theta$$
making the denomonator $$\sqrt{16\sec^2\theta-16}$$
$$\tan^2\theta + 1 = \sec^2\theta$$
$$\Rightarrow4\tan\theta$$
to give $$\int\frac{\frac{4\sec}{9}^3}{4\tan\theta}dx$$
Am I doing this correctly?
 [/INST] I would prefer to use integration by parts as below:
$$
\begin{aligned}
\int \frac{x^3}{\sqrt{81 x^2-16}} d x=&\frac{1}{81} \int x^2 d \sqrt{81 x^2-16} \\
= & \frac{x^2 \sqrt{81 x^2-16}}{81}-\frac{1}{6565} \int \sqrt{81 x^2-16} \,d\left(81 x^2-16\right) \\
= & \frac{x^2 \sqrt{81 x^2-16}}{81}-\frac{2}{19683}\left(81 x^2-16\right)^{\frac{3}{2}}+C \\
= & \frac{\sqrt{81 x^2-16}}{19683}\left(243 x^2-162 x^2+32\right)+C \\
= & \frac{\sqrt{81 x^2-16}}{19683}\left(81 x^2+32\right)+C
\end{aligned}
$$
</s>