integration by substitution

Apr 2008
204
5
\(\displaystyle \int \frac{1}{x^2 \sqrt {4+x^2}} dx\)

it says to substitute \(\displaystyle x = 2 tan \theta\)

i get down to \(\displaystyle \frac{1}{4} \int \frac{cos \theta}{1 - cos^2 \theta} d \theta\)

no idea what to do from here!
 
Dec 2009
872
381
1111
\(\displaystyle \int \frac{1}{x^2 \sqrt {4+x^2}} dx\)

it says to substitute \(\displaystyle x = 2 tan \theta\)

i get down to \(\displaystyle \frac{1}{4} \int \frac{cos \theta}{1 - cos^2 \theta} d \theta\)

no idea what to do from here!
Dear wik_chick,

\(\displaystyle \frac{1}{4} \int \frac{cos \theta}{1 - cos^2 \theta} d \theta\)

\(\displaystyle \frac{1}{4}\int\frac{cos\theta}{sin^{2}\theta}~d\theta\)

\(\displaystyle \frac{1}{4}\int{cot\theta~{cosec\theta}~d\theta}\)

\(\displaystyle -\frac{1}{4}\int\frac{d}{d\theta}cosec\theta~{d\theta}\)

Hope you can continue from here.
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
\(\displaystyle \int \frac{1}{x^2 \sqrt {4+x^2}} dx\)

it says to substitute \(\displaystyle x = 2 tan \theta\)

i get down to \(\displaystyle \frac{1}{4} \int \frac{cos \theta}{1 - cos^2 \theta} d \theta\)

no idea what to do from here!
You should find that when you make the substitutions:

Remember that \(\displaystyle 1 - \cos^2{\theta} = \sin^2{\theta}\).

Then the integral becomes

\(\displaystyle \frac{1}{4}\int{\frac{\cos{\theta}}{\sin^2{\theta}}\,d\theta}\).


Now let \(\displaystyle u = \sin{\theta}\) so that \(\displaystyle du = \cos{\theta}\,d\theta\) and the integral becomes

\(\displaystyle \frac{1}{4}\int{\frac{1}{u^2}\,du}\)

\(\displaystyle = \frac{1}{4}\int{u^{-2}\,du}\)

\(\displaystyle = -\frac{1}{4}u^{-1} + C\)

\(\displaystyle = -\frac{1}{4u} + C\)

\(\displaystyle = -\frac{1}{4\sin{\theta}} + C\).


Now remembering that \(\displaystyle \sin{\theta} = \frac{\tan{\theta}}{\sqrt{1 + \tan^2{\theta}}}\)

\(\displaystyle \sin{\theta} = \frac{\frac{x}{2}}{\sqrt{1 + \left(\frac{x}{2}\right)^2}}\)

\(\displaystyle \sin{\theta} = \frac{\frac{x}{2}}{\sqrt{\frac{4 + x^2}{4}}}\)

\(\displaystyle \sin{\theta} = \frac{\frac{x}{2}}{\frac{\sqrt{4 + x^2}}{2}}\)

\(\displaystyle \sin{\theta} = \frac{x}{\sqrt{4 + x^2}}\).


Plugging this back into the solution gives:

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

\(\displaystyle = -\frac{\sqrt{4 + x^2}}{4x} + C\).
 
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