Integrating using Trig Substitution

• Jun 14th 2008, 01:57 PM
JonathanEyoon
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. (Speechless)
• Jun 14th 2008, 02:05 PM
Mathstud28
Quote:

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. (Speechless)

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 $\varphi=\sqrt{2}\sec(\theta)$
• Jun 14th 2008, 02:33 PM
JonathanEyoon
Quote:

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?
• Jun 14th 2008, 02:35 PM
Mathstud28
Quote:

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?

$\sqrt{2\sec(\theta)^2-2}=\sqrt{2}\color{red}{\tan(\theta)}$
• Jun 14th 2008, 03:01 PM
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.
• Jun 14th 2008, 03:06 PM
Mathstud28
Quote:

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
• Jun 14th 2008, 03:16 PM
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?
• Jun 14th 2008, 03:42 PM
Mathstud28
Quote:

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

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