Trigonometric substitution

Apr 2010
156
0
Use the substitution \(\displaystyle x = 2tan\theta, -\pi/2 < \theta < \pi/2\) to evaluate:

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

Attempt:

= \(\displaystyle \int \frac{1}{4tan^2\theta\sqrt{4tan^2\theta+ 4} } dx\)

\(\displaystyle x = 2tan\theta\)
\(\displaystyle dx = 2sec^2\theta d\theta\)

= \(\displaystyle \int \frac {1}{(4tan^2\theta)|2sec\theta|}2sec^2\theta d\theta\)
=\(\displaystyle \int \frac {sec\theta}{4tan^2\theta} d\theta\)
=\(\displaystyle \frac{1}{4}\int\frac{sec\theta}{tan^2\theta}\)

I dont know what to do from here.
 
Nov 2008
1,458
646
France
Use the substitution \(\displaystyle x = 2tan\theta, -\pi/2 < \theta < \pi/2\) to evaluate:

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

Attempt:

= \(\displaystyle \int \frac{1}{4tan^2\theta\sqrt{4tan^2\theta+ 4} } dx\)

\(\displaystyle x = 2tan\theta\)
\(\displaystyle dx = 2sec^2\theta d\theta\)

= \(\displaystyle \int \frac {1}{(4tan^2\theta)|2sec\theta|}2sec^2\theta d\theta\)
=\(\displaystyle \int \frac {sec\theta}{4tan^2\theta} d\theta\)
=\(\displaystyle \frac{1}{4}\int\frac{sec\theta}{tan^2\theta}\)

I dont know what to do from here.
=\(\displaystyle \frac{1}{4}\int\frac{\cos\theta d\theta}{sin^2\theta}\)

Then use the substitution \(\displaystyle u = \sin \theta\)
 
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