# Thread: Trig Substitutions

1. ## Trig Substitutions

Let
$
x = 3\sin u
$

$
\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

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

Any help would be greatly appreciated. Thanks.

2. $\frac{1}{9}\int csc^{2}udu = -\frac{1}{9} \cot u + C$

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

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

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

Can you see how it fits ?

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

$
\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

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

Any help would be greatly appreciated. Thanks.
$
x = 3\sin u \mbox{ then } dx=2\cos(u)
$

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

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

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

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

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

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

we need to take a side trip for a moment

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

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

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

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

wow I am tired.

4. How did you know-

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

BTW, I was supposed to use a reference triangle.

5. $\cot \theta = \frac{\cos \theta}{\sin \theta}$

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

$\cot \left(\sin^{-1} x\right) = \frac{{\color{red}\cos \left(\sin^{-1} x\right)}}{{\color{blue}\sin \left(\sin^{-1} x\right)}} = \frac{{\color{red}\sqrt{1-x^{2}}}}{{\color{blue}x}}$

6. Got it, I understand now. Thanks so much!

7. A reciprocal substitution also works. (I dunno why people don't use it. )

8. Originally Posted by Krizalid
A reciprocal substitution also works. (I dunno why people don't use it. )
What is a reciporcal substitution? I might know it by a different name...but if you could explain that would be great

9. $x=\frac1u.$

10. Originally Posted by Krizalid
$x=\frac1u.$
ok...where would you go from there?

11. Go from there by yourself, you'll see.

12. Originally Posted by Krizalid
Go from there by yourself, you'll see.
$\int\frac{1}{\frac{1}{u^2}\sqrt{9-(\frac{1}{u})^2}}dx=$..........hm..........integrations by parts? $u=u^2$... $dv=\frac{1}{\sqrt{9-(\frac{1}{u})^2}}$... $du=2u$... $v=arcsin\bigg(\frac{1}{3u}\bigg)$?

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

14. That's an interesting sub o.O

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

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

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

$= -\frac{1}{3} \int \frac{du}{\sqrt{1 - \left(\frac{1}{3u}\right)^{2}}} = -\frac{1}{3} \arcsin \left(\frac{1}{3u}\right) + C$

15. It's really helpful, but not always works. In this cases you can turn the integral into an easy one.

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