1. ## trig substitution

Good afternoon,

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

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

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

I arrive at

$
{\int{\frac{\sqrt{4-x^2}}{x^2}}
=\frac{1}{2}\ln\mid\frac{2+x}{\sqrt{4-x^2}}\mid+C$

2. Hmm. I don't get your result. Could you please post a few more steps?

3. Sure,

Using the substition...

$
x=2sin\theta
$

$
dx=2cos\theta d\theta
$

I get...

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

So...

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

Wait... Should that have been...

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

This must have been where I went wrong. What should this have been, and why?

4. [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?

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

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

The substitution is good!!!

...but the final answer is wrong!

(Take the derivative of what you got, is it the integrand?)

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

...
Your numerator is indeed $2\cos \theta$ but your denominator is $x^2 = (2\sin \theta)^2$, and don't forget to replace dx with $2\cos\theta d\theta$.

7. Totally not right.

Originally Posted by MechEng
$
{\int{\frac{\sqrt{4-x^2}}{x^2}}
=\frac{1}{2}\ln\mid\frac{2+x}{\sqrt{4-x^2}}\mid+C$
[LaTeX ERROR: Convert failed]

8. ## no,

it's:

$\int cot^2\theta d\theta = -\theta- cot\theta +C$

9. 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.

it's:

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

EDIT. I see you edited your post.

10. You are blind or what??? Where are you see minus??? hahaha...

11. 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:

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

Can this be reduced to:

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

And then:

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

And then:

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

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

So:

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

And:

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

Am I totally off target again?

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

And then:

$
-\theta-cot^2\theta+c
$
Are you sure?

13. Should that look more like:

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

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

Thank you.

14. note the identity ...

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

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

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

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

back substitute ...

$\displaystyle -\frac{\sqrt{4-x^2}}{x} - \arcsin\left(\frac{x}{2}\right) + C$

15. Originally Posted by skeeter

sub in for $\cos^2{t}$ in the integrand ...
Where am I substituting this?

Originally Posted by skeeter

sub in for $\cot^2{t}$ in the integrand ...
I'm getting really hung up on my trig identities.

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