Integration using trig substitution: int 1/(4 - x^2) dx =...

[chengeto]

Use trigonometric substitution to show that:
$\displaystyle \int\frac{1}{4-x^2}dx=\frac{1}{4}ln|\frac {2+x}{2-x}|+c$

Attempt to solution:

$\displaystyle 4(1-(\frac{x}{2})^2)$

$\displaystyle \text{let} (\frac {x}{2}^2)=sin\theta$

$\displaystyle x=2sin\theta$

$\displaystyle \theta=arcsin(\frac{x}{2})$

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

$\displaystyle \int\frac{1}{4cos^2 \theta}2cos\theta d\theta$

$\displaystyle \frac{1}{2}\int sec\theta d\theta=ln|sec\theta+tan\theta|=ln|sec (arcsin\frac{x}{2})+ tan (arcsin\frac{x}{2})|$

Guys l get stuck here how do they end up with : $\displaystyle \frac{1}{4}ln|\frac {2+x}{2-x}|+c$

 

[galactus]

Re: Integration using trig substitution

That is a matter of knowing the trig inverses, such as

$\displaystyle sec(sin^{-1}(\frac{x}{2}))=\frac{2}{\sqrt{(2-x)(x+2)}}$ and $\displaystyle tan(sin^{-1}(\frac{x}{2}))=\frac{x}{\sqrt{(2-x)(x+2)}}$

$\displaystyle \frac{1}{2}ln\left(\frac{2}{\sqrt{(2-x)(x+2)}}+\frac{x}{\sqrt{(2-x)(x+2)}}\right)=\frac{1}{2}ln(\frac{x+2}{\sqrt{4-x^{2}}})$

$\displaystyle =\frac{1}{2}\left[ln(x+2)-\frac{1}{2}ln((2-x)(x+2))\right]=\frac{1}{2}\left[ln(x+2)-\frac{1}{2}ln(2-x)+\frac{1}{2}ln(2+x)\right]$

$\displaystyle =\frac{1}{4}\left[ln(x+2)-ln(2-x)\right]$

$\displaystyle =\frac{1}{4}ln(\frac{x+2}{2-x})$


Your work looked good.

 

[Deleted member 4993]

Re: Integration using trig substitution

I don't know why did your problem specifically asked for trig. substitution. Without that, the problem is much simpler:

$\displaystyle \frac{1}{4-x^2} \, \, = \, \, \frac{1}{4}\cdot [\frac{1}{2+x} \, + \, \frac{1}{2-x}]$

 

[galactus]

Re: Integration using trig substitution

Yes indeed, that's probably why they wanted trig sub.

 

[chengeto]

Re: Integration using trig substitution

I don't know why did your problem specifically asked for trig. substitution. Without that, the problem is much simpler:

$\displaystyle \frac{1}{4-x^2} \, \, = \, \, \frac{1}{4}\cdot [\frac{1}{2+x} \, + \, \frac{1}{2-x}]$

I could have used partial fraction decomposition but the question said you should only use trig substitution.

 

[chengeto]

Re: Integration using trig substitution

That is a matter of knowing the trig inverses, such as

$\displaystyle sec(sin^{-1}(\frac{x}{2}))=\frac{2}{\sqrt{(2-x)(x+2)}}$ and $\displaystyle tan(sin^{-1}(\frac{x}{2}))=\frac{x}{\sqrt{(2-x)(x+2)}}$

$\displaystyle \frac{1}{2}ln\left(\frac{2}{\sqrt{(2-x)(x+2)}}+\frac{x}{\sqrt{(2-x)(x+2)}}\right)=\frac{1}{2}ln(\frac{x+2}{\sqrt{4-x^{2}}})$

$\displaystyle =\frac{1}{2}\left[ln(x+2)-\frac{1}{2}ln((2-x)(x+2))\right]=\frac{1}{2}\left[ln(x+2)-\frac{1}{2}ln(2-x)+\frac{1}{2}ln(2+x)\right]$

$\displaystyle =\frac{1}{4}\left[ln(x+2)-ln(2-x)\right]$

$\displaystyle =\frac{1}{4}ln(\frac{x+2}{2-x})$


Your work looked good.

Thanks very much for the help but l would like to know how you got this :

$\displaystyle \frac{1}{2}\left[ln(x+2)-\frac{1}{2}ln((2-x)(x+2)\right]$

 

[galactus]

Re: Integration using trig substitution

I just factored $\displaystyle 4-x^{2}$ and used the log laws.

$\displaystyle 4-x^{2}=(2-x)(2+x)$

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

$\displaystyle ln(((2-x)(2+x))^{\frac{1}{2}})=\frac{1}{2}ln((2-x)(2+x))$