Tricky Trigonometric Integral

Question

How do we solve $\int x^3\sqrt{4+x^2}dx$? I think I substituted the wrong thing in and got the wrong answer. I've tried multiple times and cant figure it out. I got that the $\sqrt{4+x^2}= 2\sec(\theta)$. Don't know what to do from there.

Answer 1

Hints: with

$$\begin{cases}u=x^2,&u'=2x\\{}\\v'=x\sqrt{4+x^2},&v=\frac13(4+x^2)^{3/2}\end{cases}\implies\int x^3\sqrt{4+x^2}\,dx=$$$${}$$

$$=\frac13x^2(4+x^2)^{3/2}-\frac23\int x(4+x^2)^{3/2}\,dx$$

Answer 2

Hint:

use the substitution $$ 4+x^2=t $$ that gives $$ x^3dx=\frac{1}{2}(t-4)dt $$

Answer 3

$u=4+x^2\to\text{d}u=2x\text{d}x\implies \int x^3\sqrt{u}\frac{\text{d}u}{2x}$ $=\frac{1}{2}\int(4-u)u^{1/2}\text{d}u$ can you move forward from here?

Answer 4

$x=2\tan \theta$, we have \begin{align*} I=\int x^3\sqrt{4+x^2}dx&=16\int \tan^3\theta \sec \theta(1+\tan^2\theta) d\theta=16\int\frac{\sin^3\theta}{\cos^6\theta}d\theta\\&=16\int\frac{\sin\theta(1-\cos^2\theta)}{\cos^6\theta}d\theta \end{align*} set $\cos\theta=u$, therefore $$I=-16\int\frac{1-u^2}{u^6}du=\frac{16}{5u^5}-\frac{16}{3u^3}+c$$