# Evaluating $\int_0^1 \frac {x^3}{\sqrt {4+x^2}}\,dx$

How do I evaluate the definite integral $$\int_0^1 \frac {x^3}{\sqrt {4+x^2}}\,dx ?$$ I used trig substitution, and then a u substitution for $\sec\theta$.

I tried doing it and got an answer of: $-\sqrt{125}+12\sqrt{5}-16$, but apparently its wrong.

Can someone help check my error?

• It would be hard to spot the error without seeing more of your steps. Commented Jan 29, 2015 at 4:59
• Please use TeX. You can find TeX commands from here. Commented Jan 29, 2015 at 5:00
• Wow, I never knew you can use TeX. Thanks Mario for fixing it. I'll be sure to use the commands from now on. Commented Jan 29, 2015 at 5:02
• Okay @Travis , I'll post a pic from dropbox of my solution Commented Jan 29, 2015 at 5:03
• @XihaiLuo You're welcome! Commented Jan 29, 2015 at 5:03

A way to compute this is as follows: \begin{align*} \int_0^1 \frac {x^3}{\sqrt {4+x^2}}\mathrm d x &=\int_0^1\frac{4x+x^3-4x}{\sqrt{4+x^2}}\mathrm d x\\ &=\int_0^1x\sqrt{4+x^2}\mathrm d x -2\int_0^1\frac{2x}{\sqrt{4+x^2}}\mathrm d x\\ &=\left.\frac{1}{3}(4+x^2)^{3/2}\right|_0^1-4\left.\left(4+x^2\right)^{1/2}\right|_0^1\\ &=\frac{5\sqrt{5}-8}{3}-4\left(\sqrt{5}-2\right)\\ &=\boxed{\color{blue}{\dfrac{16-7\sqrt{5}}{3}}} \end{align*}

• GREAT solution, never would have thought of it! I always envy guys who are so good at basic algebra that they can solve any problem with a slight of hand like this. Commented Jan 29, 2015 at 5:21
• @Mathemagician1234 Thanks, there are some skills that one learn by observing another people. Commented Jan 29, 2015 at 5:23
• Holy moly what a solution
– qwr
Commented Jan 29, 2015 at 8:03

By the change of variable $x^2+4=t$, we have $2x\,dx=dt$: $$\int_0^1\frac {x^3}{\sqrt {4+x^2}}\,dx=\frac12\int_4^5\frac{t-4}{\sqrt t}\,dt=\frac13t\sqrt t-4\sqrt t\Big|_4^5=\frac{16-7\sqrt5}3.$$

Following your way: \begin{align}\int_0^1 \frac {x^3}{\sqrt {4+x^2}}\,dx &=8\int_0^{\arctan(1/2)}\tan^3\theta\sec\theta{\rm d}\theta\tag{x=2\tan\theta}\\&=8\int_1^{\sqrt{5}/2}(t^2-1){\rm d}t\tag{t=\sec\theta}\\&=8\left(\frac{t^3}3-t\right)\Bigg|_1^{\sqrt5/2}\\&=\frac83\left(\frac{5\sqrt5}8-1\right)-8\left(\frac{\sqrt5}2-1\right)\\&=\large \frac{16-7\sqrt5}3 \end{align}

Hint:

$$\int_0^1\frac{x^3}{\sqrt{4+x^2}}dx=\frac{1}{2}\int_0^1\frac{x^22x}{\sqrt{4+x^2}}dx=\int_0^1\frac{(x^2+4-4)(4+x^2)'}{\sqrt{4+x^2}}dx$$

Alternately, let $x=2\sinh t$, and use the fact that $\cosh^2t-\sinh^2t=1,~\sinh't=\cosh t$, and $\cosh't=\sinh t$. You will finally arrive at $\displaystyle\int\sinh^3t~dt=\int(\cosh^2t-1)~d(\cosh t)=$ $=\displaystyle\int(u^2-1)~du$, which is trivial to evaluate.

\begin{aligned} \int_0^1 \frac{x^3}{\sqrt{4+x^2}} d x&=\int_0^1 x^2 d\left(\sqrt{4+x^2}\right) \\ & =\left[x^2 \sqrt{4+x^2}\right]_0^1-\int_0^1 \sqrt{4+x^2} d\left(x^2\right) \\ & =\sqrt{5}-\left[\frac{2}{3}\left(4+x^2\right)^{\frac{3}{2}}\right]_0^1\\&=\frac{1}{3}(16-7 \sqrt{5}) \end{aligned}

Substitute $$x^2=u$$, $$du=2x dx$$ $$\int_0^1 \frac{x^3 dx}{\sqrt{4+x^2}} = \frac12 \int_0^1 \frac{u du}{\sqrt{4+u}}$$ and formula 2.222.2 in the Gradsteyn-Ryzhik tables gives $$\int \frac{u du}{\sqrt{4+u}} = \frac23 (4+u)^{3/2} -8(4+u^{1/2})$$