Evaluate $\int\frac{x^3}{\sqrt{81x^2-16}}dx$ using Trigonometric Substitution

Question

$$\int\frac{x^3}{\sqrt{81x^2-16}}dx$$ 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?

Answer 1

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

Answer 2

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

Answer 3

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

$$u=81x^2-16,du=162xdx$$ $$\int\dfrac{x^3dx}{\sqrt{81x^2-16}}=\int\dfrac{\frac1{162}x^2(162xdx)}{\sqrt{81x^2-16}}=\frac1{162}\int\dfrac{\frac{u+16}{81}du}{u^{1/2}}=$$ $$\frac1{13122}\int u^{1/2}du+\frac8{6561}\int u^{-1/2}du$$

Answer 4

I would prefer to use integration by parts as below: $$ \begin{aligned} \int \frac{x^3}{\sqrt{81 x^2-16}} d x=&\frac{1}{81} \int x^2 d \sqrt{81 x^2-16} \\ = & \frac{x^2 \sqrt{81 x^2-16}}{81}-\frac{1}{6565} \int \sqrt{81 x^2-16} \,d\left(81 x^2-16\right) \\ = & \frac{x^2 \sqrt{81 x^2-16}}{81}-\frac{2}{19683}\left(81 x^2-16\right)^{\frac{3}{2}}+C \\ = & \frac{\sqrt{81 x^2-16}}{19683}\left(243 x^2-162 x^2+32\right)+C \\ = & \frac{\sqrt{81 x^2-16}}{19683}\left(81 x^2+32\right)+C \end{aligned} $$