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Problem 7 Easy Difficulty

Use the Table of Integrals on Reference Pages 6-10 to evaluate the integral.

$ \displaystyle \int \frac{\cos x}{\sin^2 x - 9}\ dx $


$\frac{1}{6} \ln \left|\frac{\sin x-3}{\sin x+3}\right|+C$


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Video Transcript

Okay. This question wants us to use an integral table to find this anti derivative. So right now, this doesn't really look like any common forms that we see in our table. So what we have often have to do is make a clever U substitution. So a good substitution to use in this case is you equal to sign X because we see we have signs and co signs. So if we take to you, we see that that's equal to co sign of X d. X. And look at that. We have co sign of X t X. So this means are integral transforms to the integral of do you Over you squared minus nine and we can see this could be rewritten nous the form integral of d U over you squared minus a squared and based on our integral table, it says that the integral off do you over you squared minus a squared is equal to one over two times a times the natural log of you minus a over you plus a And in this case, since we have a nine in our denominator, nine is three squared, so a equals three and remember What are you? Was sine X. So now plugging back in, we get one over two times, 3 to 6 times the natural log of sign of X minus three over sign of X plus three. And if you're having trouble getting this from the integral table, you could always prove this using partial fractions.

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