# Thread: Integrating using Trig Substitution

1. ## Integrating using Trig Substitution

The problem is

(x^4) / (sqrt x^10 -2)

I thought I could use U- Substitution first and work it out like that...but that doesn't pan out. Help please this one has me stumped.

2. Originally Posted by JonathanEyoon
The problem is

(x^4) / (sqrt x^10 -2)

I thought I could use U- Substitution first and work it out like that...but that doesn't pan out. Help please this one has me stumped.
Let $x^5=\varphi$

so $d\varphi=5x^4dx$

Giving us

$\frac{-1}{5}\int\frac{d\varphi}{2-\varphi^2}$

Now let $\varphi=\sqrt{2}\tanh(\theta)$

and go from there

EDIT:

I am sorry I misread it, it should be

$\frac{1}{5}\int\frac{d\varphi}{\sqrt{\varphi^2-2}}$

Now let $\varphi=\sqrt{2}\sec(\theta)$

3. Originally Posted by Mathstud28
Let $x^5=\varphi$

so $d\varphi=5x^4dx$

Giving us

$\frac{-1}{5}\int\frac{d\varphi}{2-\varphi^2}$

Now let $\varphi=\sqrt{2}\tanh(\theta)$

and go from there

EDIT:

I am sorry I misread it, it should be

$\frac{1}{5}\int\frac{d\varphi}{\sqrt{\varphi^2-2}}$

Now let $\varphi=\sqrt{2}\sec(\theta)$

OK I worked it out from where you left off and eventually I get to a point where it looks like this

(1/5) integrate (SecTheta)

When i integrate that, I end up with

(1/5) ln(abs(secTheta + tan Theta))

What am i supposed to do from here?

4. Originally Posted by JonathanEyoon
OK I worked it out from where you left off and eventually I get to a point where it looks like this

(1/5) integrate (SecTheta)

When i integrate that, I end up with

(1/5) ln(abs(secTheta + tan Theta))

What am i supposed to do from here?
$\sqrt{2\sec(\theta)^2-2}=\sqrt{2}\color{red}{\tan(\theta)}$

5. Ok made the correction for tanTheta and ended up with having to integrate secTheta which my final line is

(1/5) ln l sectheta + tantheta l + C

Now looking at my teachers answer, this is far from correct but i'm not sure as to how to proceed from here.

6. Originally Posted by JonathanEyoon
Ok made the correction for tanTheta and ended up with having to integrate secTheta which my final line is

(1/5) ln l sectheta + tantheta l + C

Now looking at my teachers answer, this is far from correct but i'm not sure as to how to proceed from here.
$\frac{1}{5\sqrt{2}}\int\sec(\theta)d\theta=\frac{1 }{5\sqrt{2}}\ln|\sec(\theta)+\tan(\theta)|$

$x^5=\varphi=\sqrt{2}\sec(\theta)\Rightarrow{arcsec \bigg(\frac{x^5}{\sqrt{2}}
\bigg)=\theta}$

So now back sub

7. MmMm.... how are you getting those values? I remember the teacher saying something about drawing a triangle to figure out how to get everything back in terms of x. I think this is the part but I didn't understand it too well. Could you show me?

8. Originally Posted by JonathanEyoon
MmMm.... how are you getting those values? I remember the teacher saying something about drawing a triangle to figure out how to get everything back in terms of x. I think this is the part but I didn't understand it too well. Could you show me?
Obviously

$\sec\bigg(arcsec\bigg(\frac{x^5}{\sqrt{2}}\bigg)\b igg)=\frac{x^5}{\sqrt{2}}$

and

$\tan\bigg(arcsec\bigg(\frac{x^5}{\sqrt{2}}\bigg)\b igg)=\frac{\sqrt{x^{10}-2}}{\sqrt{2}}$