$\int\frac{x^3}{\sqrt{4+x^2}}$

Question

I was trying to calculate

$$\int\frac{x^3}{\sqrt{4+x^2}}$$

Doing $x = 2\tan(\theta)$, $dx = 2\sec^2(\theta)~d\theta$, $-\pi/2 < 0 < \pi/2$ I have:

$$\int\frac{\left(2\tan(\theta)\right)^3\cdot2\cdot\sec^2(\theta)~d\theta}{2\sec(\theta)}$$

which is

$$8\int\tan(\theta)\cdot\tan^2(\theta)\cdot\sec(\theta)~d\theta$$

now I got stuck ... any clues what's the next substitution to do? I'm sorry for the formatting. Could someone please help me with the formatting?

Answer 1

You have not chosen an efficient way to proceed. However, let us continue along that path.

Note that $\tan^2\theta=\sec^2\theta-1$. So you want $$\int 8(\sec^2\theta -1)\sec\theta\tan\theta\,d\theta.$$ Let $u=\sec\theta$.

Remark: My favourite substitution for this problem and close relatives is a variant of the one used by Ayman Hourieh. Let $x^2+4=u^2$. Then $2x\,dx=2u\,du$, and $x^2=u^2-4$. So $$\int \frac{x^3}{\sqrt{x^2+4}}\,dx=\int \frac{(u^2-4)u}{u}\,du=\int (u^2-4)\,du.$$

Answer 2

Let $u = x^2 + 4$, $du = 2x\,dx$:

\begin{align*} I &= \frac{1}{2} \int \frac{u - 4}{\sqrt{u}}du \end{align*}

Should be easy to take it from there.

Answer 3

HINT: $\tan^2\theta=\sec^2\theta-1$, and $d(\sec\theta)=\sec\theta\tan\theta~d\theta$.

Answer 4

$$ \begin{aligned} \int \frac{x^{3}}{\sqrt{4+x^{2}}} d x &=\int x^{2} d \sqrt{4+x^{2}} \\ &\stackrel{IBP}{=} x^{2} \sqrt{4+x^{2}}-\int \sqrt{4+x^{2}} d\left(x^{2}\right) \\ &=x^{2} \sqrt{4+x^{2}}-\frac{2}{3}\left(4+x^{2}\right)^{\frac{3}{2}}+C \\ &=\frac{\sqrt{4+x^{2}}}{3}\left[3 x^{2}-2\left(4+x^{2}\right)\right]+C \\ &=\frac{\sqrt{4+x^{2}}}{3}\left(x^{2}-8\right)+C \end{aligned} $$