# trig substitution

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• Jun 25th 2010, 10:45 AM
MechEng
trig substitution
Good afternoon,

I have worked out the following problem and arrived at the following answer. Would someone mind verifying this for me?

$\displaystyle {\int{\frac{\sqrt{4-x^2}}{x^2}}dx$

let $\displaystyle x=2sin\theta$
let $\displaystyle dx=2cos\theta d\theta$

I arrive at

$\displaystyle {\int{\frac{\sqrt{4-x^2}}{x^2}} =\frac{1}{2}\ln\mid\frac{2+x}{\sqrt{4-x^2}}\mid+C$
• Jun 25th 2010, 11:06 AM
Ackbeet
Hmm. I don't get your result. Could you please post a few more steps?
• Jun 25th 2010, 11:36 AM
MechEng
Sure,

Using the substition...

$\displaystyle x=2sin\theta$
$\displaystyle dx=2cos\theta d\theta$

I get...

$\displaystyle \sqrt{4-x^2} = \sqrt{4-4sin^2\theta} = 2\sqrt{1-sin^2\theta} = 2\sqrt{cos^2\theta} = 2cos\theta$

So...

$\displaystyle \int{\frac{2cos\theta}{4cos^2\theta}d\theta}$

Wait... Should that have been...

$\displaystyle \int{\frac{d\theta}{4cos^2\theta}}$

This must have been where I went wrong. What should this have been, and why?
• Jun 25th 2010, 11:37 AM
Also sprach Zarathustra
[QUOTE=MechEng;530685]Good afternoon,

I have worked out the following problem and arrived at the following answer. Would someone mind verifying this for me?

$\displaystyle {\int{\frac{\sqrt{4-x^2}}{x^2}}dx$

let $\displaystyle x=2sin\theta$
let $\displaystyle dx=2cos\theta d\theta$

The substitution is good!!!

...but the final answer is wrong!

(Take the derivative of what you got, is it the integrand?)
• Jun 25th 2010, 11:38 AM
Ackbeet
• Jun 25th 2010, 11:45 AM
slider142
Quote:

Originally Posted by MechEng
...
$\displaystyle \int{\frac{2cos\theta}{4cos^2\theta}d\theta}$
...

Your numerator is indeed $\displaystyle 2\cos \theta$ but your denominator is $\displaystyle x^2 = (2\sin \theta)^2$, and don't forget to replace dx with $\displaystyle 2\cos\theta d\theta$.
• Jun 25th 2010, 11:47 AM
TheCoffeeMachine
Totally not right.

Quote:

Originally Posted by MechEng
$\displaystyle {\int{\frac{\sqrt{4-x^2}}{x^2}} =\frac{1}{2}\ln\mid\frac{2+x}{\sqrt{4-x^2}}\mid+C$

$\displaystyle \dfrac{1}{2}\ln|\dfrac{2+x}{\sqrt{4-x^2}}| = \dfrac{1}{2}\text{arctanh}\dfrac{x}{2}\Rightarrow \bigg\{\dfrac{1}{2}\ln|\dfrac{2+x}{\sqrt{4-x^2}}|\bigg\}' = \dfrac{1}{1-4x^2}.$
• Jun 25th 2010, 11:47 AM
Also sprach Zarathustra
no,
it's:

$\displaystyle \int cot^2\theta d\theta = -\theta- cot\theta +C$
• Jun 25th 2010, 11:53 AM
TheCoffeeMachine
Quote:

Originally Posted by Also sprach Zarathustra
No

What? The OP has not yet responded. I don't know what on Earth you are referring to.

Quote:

it's:

$\displaystyle \int cot^2\theta d\theta = -\theta + cot\theta +C$
The plus should be minus.

EDIT. I see you edited your post.
• Jun 25th 2010, 12:08 PM
Also sprach Zarathustra
You are blind or what??? Where are you see minus??? hahaha...
• Jun 29th 2010, 11:15 AM
MechEng
Good afternoon All,

I am sitting at work without my notes today, trying to pickup where I left off on this one.

Using the "correct" substitution I get the following:

$\displaystyle \int{\frac{2cos\theta}{(2sin\theta)^2}(2cos\theta) d\theta$

Can this be reduced to:

$\displaystyle \int{\frac{(2cos\theta)^2}{(2sin\theta)^2}d\theta$

And then:

$\displaystyle \int{cot^2\theta}d\theta$

And then:

$\displaystyle -\theta-cot^2\theta+c$

Since $\displaystyle x=2sin\theta$ we have $\displaystyle cot\theta=\frac{\sqrt{4-x^2}}{x}$ and $\displaystyle \theta=sin^{-1}(\frac{x}{2})$

So:

$\displaystyle -sin^{-1}(\frac{x}{2})-(\frac{\sqrt{4-x^2}}{x})^2+C$

And:

$\displaystyle -sin^{-1}(\frac{x}{2})-\frac{\mid4-x^2\mid}{x^2}+C$

Am I totally off target again?
• Jun 29th 2010, 11:37 AM
TheCoffeeMachine
Quote:

Originally Posted by MechEng
$\displaystyle \int{cot^2\theta}d\theta$

And then:

$\displaystyle -\theta-cot^2\theta+c$

Are you sure?
• Jun 29th 2010, 11:58 AM
MechEng
Should that look more like:

$\displaystyle \int{cot^2\theta}d\theta$ = $\displaystyle \ln{\mid sin^2\theta\mid+C$

I guess I'm confused by the $\displaystyle cot^2\theta$. Well, that and I am trying to do this without notes of any kind... between phone calls and e-mails.

Thank you.
• Jun 29th 2010, 12:15 PM
skeeter
note the identity ...

$\displaystyle \displaystyle \cot^2{t} = \csc^2{t} - 1$

sub in for $\displaystyle \cot^2{t}$ in the integrand ...

$\displaystyle \displaystyle \int \csc^2{t} - 1 \, dt = -\cot{t} - t + C$

$\displaystyle \displaystyle t = \arcsin\left(\frac{x}{2}\right)$

back substitute ...

$\displaystyle \displaystyle -\frac{\sqrt{4-x^2}}{x} - \arcsin\left(\frac{x}{2}\right) + C$
• Jun 29th 2010, 12:27 PM
MechEng
Quote:

Originally Posted by skeeter

sub in for $\displaystyle \cos^2{t}$ in the integrand ...

Where am I substituting this?

sub in for $\displaystyle \cot^2{t}$ in the integrand ...