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?

Solution 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

Solution 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