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I have to integrate $\frac{1}{1-x^2}dx$ using trig substitution. I know there is a way to do it with partial fractions which yields an answer of $\frac{-1}{2}\ln|1-x|+\frac{1}{2}\ln|1+x|+C$, however I have to show that the trig substitution and partial fractions methods both work.

I know for this problem that $x=\sin\theta, dx=\cos\theta\ d\theta, x^2=\sin^2\theta.$ So now I have $$\int \frac{\cos\theta}{1-\sin^2\theta} d\theta = \int \sec\theta \ d\theta = \ln|sec\theta+\tan\theta|.$$ This is where I have my problem because I don't know what to substitute back in for $\sec\theta$ and $\tan\theta.$

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HINT

If you have $ \sin(\theta) = x $, $ \cos (\theta) = \sqrt{1-x^2} $.

Now, $ \tan(\theta) = \dfrac{\sin(\theta)}{\cos(\theta)} $ and $ \sec(\theta) = \dfrac{1}{\cos(\theta)} $

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  • $\begingroup$ So I got ln|(1-x^2)^(1/2)+x/(1-x^2)^(1/2)|. Should I get a common denominator and get ln|(1-x^2+x)/(1-x^2)^(1/2)|. And then should I simplify this by using rules for natural logs in order to get it equal to the same thing I had for the partial fractions solution? $\endgroup$ – mmm Sep 18 '14 at 1:22
  • $\begingroup$ @mmm How did you get $ \sec(\theta) = (1-x^2)^{1/2} $? $\endgroup$ – hjpotter92 Sep 18 '14 at 1:25
  • $\begingroup$ Oh, I see I should have had 1/(1-x^2)^(1/2) shouldn't I? $\endgroup$ – mmm Sep 18 '14 at 1:27
  • $\begingroup$ @mmm Yes. Then you'll have $ \sec + \tan = \frac{1+\sin}{\cos} $. $\endgroup$ – hjpotter92 Sep 18 '14 at 1:28
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$$\sec \theta = \frac{1}{\cos \theta} = \frac{1}{\sqrt{1 - \sin^2 \theta}} \\ \tan \theta = \sqrt{\sec^2 \theta - 1} = \sqrt{\frac{1-\cos^2 \theta}{\cos^2 \theta}} = \sqrt{\frac{\sin^2 \theta}{1-\sin^2\theta}}$$

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