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?
 A: If Trigonometric Substitution is not mandatory,
make the following substitution  $$\sqrt{81x^2-16}=u$$
$$\implies81x^2-16=u^2\implies162x\ dx=2u\ du\iff x\ dx=\frac{u\ du}{81}$$
and $x^2=\dfrac{u^2+16}{81}$
Hope you can take it from here
A: Set $9x=4\sec\theta\implies81x^2-16=(4\tan\theta)^2$
$$\int\frac{x^3}{\sqrt{81x^2-16}}dx$$
$$=\int\frac{4^3\sec^3\theta}{9^3\cdot4\tan\theta\cdot(\text{sign of}\tan\theta)}\frac49\sec\theta\tan\theta\ d\theta$$
$$=\frac{4^3}{9^4\cdot(\text{sign of}\tan\theta)}\int(1+\tan^2\theta)\sec^2\theta\ d\theta$$
Hope you can take it home from here
A: First, to answer the question as asked, you forgot about converting from $du$ to $dx$, then from $du$ to $d\theta$.  For the first substitution,
$$u=9x,du=9dx,dx=\frac19du$$
So after that step, you should take what you currently had after the first step and divide it by $9$.  For the second substitution
$$u=4\sec\theta,du=4\sec\theta\tan\theta d\theta$$
This should leave you with
$$\int\dfrac{\dfrac{4^3\sec^3\theta}{9^4}}{4\tan\theta}(4\sec\theta\tan\theta d\theta)=\int\frac{4^3}{9^4}(\tan^2\theta+1)\sec^2\theta d\theta$$
As an alternate substitution (trigonometric substitution is not always the best way), you could use
$$u=81x^2-16,du=162xdx$$
$$\int\dfrac{x^3dx}{\sqrt{81x^2-16}}=\int\dfrac{\frac1{162}x^2(162xdx)}{\sqrt{81x^2-16}}=\frac1{162}\int\dfrac{\frac{u+16}{81}du}{u^{1/2}}=$$
$$\frac1{13122}\int u^{1/2}du+\frac8{6561}\int u^{-1/2}du$$
A: I would prefer to use integration by parts as below:
$$
\begin{aligned}
\int \frac{x^3}{\sqrt{81 x^2-16}} d x=&\frac{1}{81} \int x^2 d \sqrt{81 x^2-16} \\
= & \frac{x^2 \sqrt{81 x^2-16}}{81}-\frac{1}{6565} \int \sqrt{81 x^2-16} \,d\left(81 x^2-16\right) \\
= & \frac{x^2 \sqrt{81 x^2-16}}{81}-\frac{2}{19683}\left(81 x^2-16\right)^{\frac{3}{2}}+C \\
= & \frac{\sqrt{81 x^2-16}}{19683}\left(243 x^2-162 x^2+32\right)+C \\
= & \frac{\sqrt{81 x^2-16}}{19683}\left(81 x^2+32\right)+C
\end{aligned}
$$
