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

Question

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. — Travis Willse, Jan 29 '15 at 04:59
Wow, I never knew you can use TeX. Thanks Mario for fixing it. I'll be sure to use the commands from now on. — Xihai Luo, Jan 29 '15 at 05:02
https://www.dropbox.com/s/tkspdlwb1f6i9hr/20150129_000328.jpg?dl=0 — Xihai Luo, Jan 29 '15 at 05:05
@XihaiLuo I can't quite read your handwriting,but it's clearly an ugly computation by hand. — Mathemagician1234, Jan 29 '15 at 05:09
@XihaiLuo, See http://math.stackexchange.com/questions/1124550/what-is-the-integral-of-frac-sqrtx2-4xdx — lab bhattacharjee, Jan 29 '15 at 05:16
@labbhattacharjee That's a somewhat different integral,but the same basic substitutions.Worth a look for the user. — Mathemagician1234, Jan 29 '15 at 05:18
@Mathemagician1234, Whenever, we have $$\int x^{2n+1}(\sqrt{a^2+x^2})^m\ dx$$ we can try this — lab bhattacharjee, Jan 29 '15 at 05:20

score 7 · Answer 1 · answered Jan 29 '15 at 05:14

7

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*}

answered Jan 29 '15 at 05:14

Ángel Mario Gallegos

19,882

1

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. – Mathemagician1234 Jan 29 '15 at 05:21
@Mathemagician1234 Thanks, there are some skills that one learn by observing another people. – Ángel Mario Gallegos Jan 29 '15 at 05:23
Holy moly what a solution – qwr Jan 29 '15 at 08:03

score 3 · Answer 2 · answered Jan 29 '15 at 08:08

3

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

answered Jan 29 '15 at 08:08

score 1 · Answer 3 · answered Jan 29 '15 at 07:20

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

score 1 · Answer 4 · answered Jan 29 '15 at 08:56

1

Hint:

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

answered Jan 29 '15 at 08:56

mnsh

5,875

score 1 · Answer 5 · answered Jan 29 '15 at 10:41

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.

score 1 · Answer 6 · answered Jun 16 '23 at 05:03

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

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

6 Answers6