1. ## Trigonometric Substitution Problem

I'm trying to solve the following integral:
{ stands for integral

{x^3/(sqrt[x^2 +4]) dx

This is what I have so far:

@ stands for pheta

x=2tan@
dx=2sec^2@ d@

x^2+4=4(1+tan^2@)
Sqrt[x^2+4]=2sec@

Plugging it back in:

{8tan^3@2sec^2@/2sec@ d@

which simplifies to

8{tan^3@sec@ d@

I can't seem to solve this. I've tried integration by parts as well as simple sub.

Where am I going wrong?

Thank you!!

2. Originally Posted by Jeffman50
I'm trying to solve the following integral:
{ stands for integral

{x^3/(sqrt[x^2 +4]) dx

This is what I have so far:

@ stands for pheta

x=2tan@
dx=2sec^2@ d@

x^2+4=4(1+tan^2@)
Sqrt[x^2+4]=2sec@

Plugging it back in:

{8tan^3@2sec^2@/2sec@ d@

which simplifies to

8{tan^3@sec@ d@

I can't seem to solve this. I've tried integration by parts as well as simple sub.

Where am I going wrong?

Thank you!!
Use a u sub

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

Let $u=x^2+4 \iff x^2=u-4 \implies du=2xdx$

subbing into the above integral gives

$\frac{1}{2}\int \frac{u-4}{\sqrt{u}}du=\frac{1}{2}\int (u^{\frac{1}{2}}-4u^{-\frac{1}{2}})du$

3. Interesting.. Is there a way to do it using trigonometric substitution? Its under that heading in my textbook

4. Alternatively,

$8\int \text{tan}^3\theta \, \text{sec}\theta \, d\theta=8\int \text{tan}^2\theta \, \text{sec}\theta\text{tan}\theta \, d\theta = 8\int \left(\text{sec}^2\theta-1\right)\text{sec}\theta\text{tan}\theta \, d\theta$.

Let $u=\text{sec}\theta$.

5. Originally Posted by Jeffman50
I'm trying to solve the following integral:
{ stands for integral

{x^3/(sqrt[x^2 +4]) dx

This is what I have so far:

@ stands for pheta

x=2tan@
dx=2sec^2@ d@

x^2+4=4(1+tan^2@)
Sqrt[x^2+4]=2sec@

Plugging it back in:

{8tan^3@2sec^2@/2sec@ d@

which simplifies to

8{tan^3@sec@ d@

I can't seem to solve this. I've tried integration by parts as well as simple sub.

Where am I going wrong?

Thank you!!
You're doing great.

$\frac{d}{d\theta}(\sec\theta)=\sec\theta\tan\theta$

$\tan^2\theta=\sec^2\theta-1$

Rewrite $8\tan^3\theta\sec\theta\,d\theta$ as

$8\tan^2\theta\sec\theta\tan\theta\,d\theta$

Good luck

6. Originally Posted by TheEmptySet
Use a u sub

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

Let $u=x^2+4 \iff x^2=u-4 \implies du=2xdx$

subbing into the above integral gives

$\frac{1}{2}\int \frac{u-4}{\sqrt{u}}du=\frac{1}{2}\int (u^{\frac{1}{2}}-4u^{-\frac{1}{2}})du$
Your numerator is $x^3$, not $x^2$.

Trigonometric substitution is possible, but difficult.

It's easier to use Hyperbolic Substitution in this case.

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

Make the substitution $x = 2\sinh{t}$ so that $dx = 2\cosh{t}\,dt$.

Note that $t = \sinh^{-1}{\frac{x}{2}}$.

So the integral becomes

$\int{\frac{(2\sinh{t})^3}{\sqrt{(2\sinh{t})^2 + 4}}\,2\cosh{t}\,dt}$

$= \int{\frac{16\sinh^3{t}\cosh{t}}{\sqrt{4\sinh^2{t} + 4}}\,dt}$

$= \int{\frac{16\sinh^3{t}\cosh{t}}{\sqrt{4(\sinh^2{t } + 1)}}\,dt}$

$= \int{\frac{16\sinh^3{t}\cosh{t}}{2\sqrt{\cosh^2{t} }}\,dt}$

$= \int{\frac{8\sinh^3{t}\cosh{t}}{\cosh{t}}\,dt}$

$= 8\int{\sinh^3{t}\,dt}$

$= 8\int{\sinh{t}\sinh^2{t}\,dt}$

$= 8\int{\sinh{t}(\cosh^2{t} - 1)\,dt}$.

Now let $u = \cosh{t}$ so that $du = \sinh{t}\,dt$.

The integral becomes

$8\int{u^2 - 1\,du}$

$= 8\left[\frac{1}{3}u^3 - u\right] + C$

$= 8\left[\frac{1}{3}\cosh^3{t} - \cosh{t}\right] + C$

$= 8\left[\frac{1}{3}\sqrt{(1 + \sinh^2{t})^3} - \sqrt{1 + \sinh^2{t}}\right] + C$

$= 8\left\{\frac{1}{3}\left[\sqrt{1 + \left(\frac{x}{2}\right)^2}\right]^3 - \sqrt{1 + \left(\frac{x}{2}\right)^2}\right\} + C$

$= 8\left\{\frac{1}{3}\left[\frac{\sqrt{4 + x^2}}{2}\right]^3 - \frac{\sqrt{4 + x^2}}{2}\right\} + C$

$= 8\left[\frac{\sqrt{(4 + x^2)^3}}{24} - \frac{\sqrt{4 + x^2}}{2}\right] + C$

$= \frac{\sqrt{(4 + x^2)^3}}{3} - 4\sqrt{4 + x^2} + C$