# Trig Substitutions

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• Apr 20th 2008, 08:04 PM
kid funky fried
Trig Substitutions
Let
$\displaystyle x = 3\sin u$

$\displaystyle \int {\frac{{dx}} {{x^2 \sqrt {9 - x^2 } }}} = \frac{1} {9}\int {\csc ^2 udu} = - \frac{1} {9}(\frac{{\sqrt {9 - x^2 } }} {x}) + C$

I understand everything until the end. Specifically,
how to get

$\displaystyle (\frac{{\sqrt {9 - x^2 } }} {x})$

Any help would be greatly appreciated. Thanks.
• Apr 20th 2008, 08:16 PM
o_O
$\displaystyle \frac{1}{9}\int csc^{2}udu = -\frac{1}{9} \cot u + C$

$\displaystyle x = 3\sin u \quad \Rightarrow \quad u = \sin^{-1}\left(\frac{x}{3}\right)$

$\displaystyle \int \frac{dx}{x^{2}\sqrt{9-x^{2}}} = -\frac{1}{9} \cot\left(\sin^{-1} \left(\frac{x}{3}\right)\right) + C$

Note that: $\displaystyle \cot \left(\sin^{-1} x\right) = \frac{\sqrt{1-x^{2}}}{x}$

Can you see how it fits ?
• Apr 20th 2008, 08:22 PM
TheEmptySet
Quote:

Originally Posted by kid funky fried
Let
$\displaystyle x = 3\sin u$

$\displaystyle \int {\frac{{dx}} {{x^2 \sqrt {9 - x^2 } }}} = \frac{1} {9}\int {\csc ^2 udu} = - \frac{1} {9}(\frac{{\sqrt {9 - x^2 } }} {x}) + C$

I understand everything until the end. Specifically,
how to get

$\displaystyle (\frac{{\sqrt {9 - x^2 } }} {x})$

Any help would be greatly appreciated. Thanks.

$\displaystyle x = 3\sin u \mbox{ then } dx=2\cos(u)$

$\displaystyle \int {\frac{{dx}}{{x^2 \sqrt {9 - x^2 } }}}$

$\displaystyle \int \frac{3\cos(u)}{9\sin^2(u)\sqrt{9-9\sin^2{u}}}du$

$\displaystyle \frac{1}{9}\int \frac{\cos(u)}{\sin^2(u)\sqrt{1-\sin^2{u}}}du$

$\displaystyle \frac{1}{9}\int \frac{\cos(u)}{\sin^2(u)\sqrt{\cos^2{u}}}du$

$\displaystyle \frac{1}{9}\int \frac{1}{\sin^2(u)}du$

$\displaystyle \frac{1}{9}\int \csc^{u}du=-\frac{1}{9}\cot{u}+c$

we need to take a side trip for a moment

$\displaystyle x = 3\sin u \iff \frac{x}{3}=\sin(u) \iff \left( \frac{x}{3}\right)=\sin^{2}u \iff \left( \frac{x}{3}\right)^2= 1-\cos^2(u)$

$\displaystyle \cos^2(u)=1-\left( \frac{x}{3}\right)^2 \iff \cos(u)=\sqrt{1-\left( \frac{x}{3}\right)^2}=\frac{1}{3}\sqrt{9-x^2}$

$\displaystyle =-\frac{1}{9}\cot{u}+c=-\frac{1}{9}\frac{\cos(u)}{\sin(u)}+c=-\frac{1}{9}\frac{\frac{1}{3}\sqrt{9-x^2}}{\frac{x}{3}}+c$

$\displaystyle -\frac{1}{9}\frac{\sqrt{9-x^2}}{x}+c$

wow I am tired. (Rock)
• Apr 20th 2008, 08:27 PM
kid funky fried
How did you know-

$\displaystyle \cot (\sin ^{ - 1} x) = \frac{{\sqrt {1 - x^2 } }} {x}$

BTW, I was supposed to use a reference triangle.
• Apr 20th 2008, 08:32 PM
o_O
$\displaystyle \cot \theta = \frac{\cos \theta}{\sin \theta}$

$\displaystyle \theta = \sin^{-1} x$

$\displaystyle \cot \left(\sin^{-1} x\right) = \frac{{\color{red}\cos \left(\sin^{-1} x\right)}}{{\color{blue}\sin \left(\sin^{-1} x\right)}} = \frac{{\color{red}\sqrt{1-x^{2}}}}{{\color{blue}x}}$
• Apr 20th 2008, 08:32 PM
kid funky fried
Got it, I understand now. Thanks so much!
• Apr 21st 2008, 10:46 AM
Krizalid
A reciprocal substitution also works. (I dunno why people don't use it. :D)
• Apr 21st 2008, 12:38 PM
Mathstud28
Quote:

Originally Posted by Krizalid
A reciprocal substitution also works. (I dunno why people don't use it. :D)

What is a reciporcal substitution? I might know it by a different name...but if you could explain that would be great(Hi)
• Apr 21st 2008, 02:17 PM
Krizalid
$\displaystyle x=\frac1u.$
• Apr 21st 2008, 02:19 PM
Mathstud28
Quote:

Originally Posted by Krizalid
$\displaystyle x=\frac1u.$

ok...where would you go from there?
• Apr 21st 2008, 02:20 PM
Krizalid
Go from there by yourself, you'll see.
• Apr 21st 2008, 02:28 PM
Mathstud28
Quote:

Originally Posted by Krizalid
Go from there by yourself, you'll see.

$\displaystyle \int\frac{1}{\frac{1}{u^2}\sqrt{9-(\frac{1}{u})^2}}dx=$..........hm..........integrations by parts? $\displaystyle u=u^2$...$\displaystyle dv=\frac{1}{\sqrt{9-(\frac{1}{u})^2}}$...$\displaystyle du=2u$...$\displaystyle v=arcsin\bigg(\frac{1}{3u}\bigg)$?
• Apr 21st 2008, 02:33 PM
Mathstud28
Quote:

Originally Posted by Mathstud28
$\displaystyle \int\frac{1}{\frac{1}{u^2}\sqrt{9-(\frac{1}{u})^2}}dx=$..........hm..........integrations by parts? $\displaystyle u=u^2$...$\displaystyle dv=\frac{1}{\sqrt{9-(\frac{1}{u})^2}}$...$\displaystyle du=2u$...$\displaystyle v=arcsin\bigg(\frac{1}{3u}\bigg)$?

:D...I forgot dx
• Apr 21st 2008, 02:35 PM
o_O
That's an interesting sub o.O

$\displaystyle x = \frac{1}{u}$
$\displaystyle dx = - \frac{1}{u^{2}}du$

$\displaystyle \int \frac{-\frac{1}{u^{2}}du}{\frac{1}{u^{2}} \sqrt{9 - \frac{1}{u^{2}}}}$

$\displaystyle = - \int \frac{du}{\sqrt{9 - \frac{1}{u^{2}}}}$

$\displaystyle = -\frac{1}{3} \int \frac{du}{\sqrt{1 - \left(\frac{1}{3u}\right)^{2}}} = -\frac{1}{3} \arcsin \left(\frac{1}{3u}\right) + C$
• Apr 21st 2008, 02:36 PM
Krizalid
It's really helpful, but not always works. In this cases you can turn the integral into an easy one.
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