Trig substitution

**ur5pointos2slo** · April 5th 2009, 09:38 AM

$ \int\frac{dx}{4+x^2}\ $

I am supposed to use trig substitution here. I used a tangent sub but then everything seems to cancel itself out. Which sub would I need to use here?

**galactus** · April 5th 2009, 09:45 AM

You're correct, use $x=2tan{\theta}, \;\ dx=2sec^{2}{\theta}d{\theta}$

$\int\frac{2sec^{2}{\theta}}{4+4tan^{2}{\theta}}d{\ theta}$

$\int\frac{2sec^{2}{\theta}}{4(1+tan^{2}{\theta})}d {\theta}$

$\frac{1}{4}\int\frac{2sec^{2}{\theta}}{sec^{2}{\th eta}}d{\theta}$

Yes, the $sec^{2}{\theta}$ does cancel. Which makes it easy.

$\frac{1}{4}\int 2d{\theta}$

$\frac{1}{2}\int d{\theta}$

$\frac{\theta}{2}$

But, from our sub, ${\theta}=tan^{-1}(\frac{x}{2})$

So, we get:

$\frac{tan^{-1}(\frac{x}{2})}{2}$

**Jhevon** · April 5th 2009, 09:45 AM

Originally Posted by ur5pointos2slo

$ \int\frac{dx}{4+x^2}\ $

I am supposed to use trig substitution here. I used a tangent sub but then everything seems to cancel itself out. Which sub would I need to use here?

there is a straight forward formula for this, but if you want to do the trig sub, that's fine. you have an expression of the form $x^2 + a^2$ thus your substitution must be $x = a \tan \theta$ . that is, $x = 2 \tan \theta$

**skeeter** · April 5th 2009, 09:47 AM

Originally Posted by ur5pointos2slo

$ \int\frac{dx}{4+x^2}\ $

I am supposed to use trig substitution here. I used a tangent sub but then everything seems to cancel itself out. Which sub would I need to use here?

$x = 2\tan{t}$

$dx = 2\sec^2{t} \, dt$

$\int \frac{1}{4 + x^2} \, dx$

$\int \frac{1}{4(1 + \tan^2{t})} \, 2\sec^2{t} \, dt$

$\frac{1}{2}\int \, dt$

$\frac{1}{2} t + C$

$\frac{1}{2} \arctan\left(\frac{x}{2}\right) + C$

**ur5pointos2slo** · April 5th 2009, 09:47 AM

Originally Posted by Jhevon

there is a straight forward formula for this, but if you want to do the trig sub, that's fine. you have an expression of the form $x^2 + a^2$ thus your substitution must be $x = a \tan \theta$ . that is, $x = 2 \tan \theta$

I couldn't think of a formula that this integral could fit. Which formula are you referring to?

**ur5pointos2slo** · April 5th 2009, 09:49 AM

Originally Posted by galactus

You're correct, use $x=2tan{\theta}, \;\ dx=2sec^{2}{\theta}d{\theta}$

$\int\frac{2sec^{2}{\theta}}{4+4tan^{2}{\theta}}d{\ theta}$

$\int\frac{2sec^{2}{\theta}}{4(1+tan^{2}{\theta})}d {\theta}$

$\frac{1}{4}\int\frac{2sec^{2}{\theta}}{sec^{2}{\th eta}}d{\theta}$

Yes, the $sec^{2}{\theta}$ does cancel. Which makes it easy.

$\frac{1}{4}\int 2d{\theta}$

$\frac{1}{2}\int d{\theta}$

$\frac{\theta}{2}$

But, from our sub, ${\theta}=tan^{-1}(\frac{x}{2})$

So, we get:

$\frac{tan^{-1}(\frac{x}{2})}{2}$

I forgot all about the dtheta in there. I ended up with just the 1/2. Thanks a lot

**Jhevon** · April 5th 2009, 09:51 AM

Originally Posted by ur5pointos2slo

I couldn't think of a formula that this integral could fit. Which formula are you referring to?

on the inside cover of your calc text, there should be a formula that looks a little like this: $\int \frac {dx}{a^2 + x^2} = \frac 1a \tan^{-1} \frac xa + C$

**ur5pointos2slo** · April 5th 2009, 09:58 AM

Originally Posted by Jhevon

on the inside cover of your calc text, there should be a formula that looks a little like this: $\int \frac {dx}{a^2 + x^2} = \frac 1a \tan^{-1} \frac xa + C$

O yes I totally forgot all about that. thanks for the reminder!

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