0
$\begingroup$

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 $\tan\theta.$

$\endgroup$

2 Answers 2

0
$\begingroup$

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)} $

$\endgroup$
4
  • $\begingroup$ 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, 2014 at 1:22
  • $\begingroup$ @mmm How did you get $ \sec(\theta) = (1-x^2)^{1/2} $? $\endgroup$
    – hjpotter92
    Sep 18, 2014 at 1:25
  • $\begingroup$ Oh, I see I should have had 1/(1-x^2)^(1/2) shouldn't I? $\endgroup$
    – mmm
    Sep 18, 2014 at 1:27
  • $\begingroup$ @mmm Yes. Then you'll have $ \sec + \tan = \frac{1+\sin}{\cos} $. $\endgroup$
    – hjpotter92
    Sep 18, 2014 at 1:28
0
$\begingroup$

$$\sec \theta = \frac{1}{\cos \theta} = \frac{1}{\sqrt{1 - \sin^2 \theta}} \\ \tan \theta = \sqrt{\sec^2 \theta - 1} = \sqrt{\frac{1-\cos^2 \theta}{\cos^2 \theta}} = \sqrt{\frac{\sin^2 \theta}{1-\sin^2\theta}}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.