# Video: APCALC02AB-P1A-Q04-754150608179

Calculate β« cscΒ²3π₯ dπ₯.

03:40

### Video Transcript

Calculate the indefinite integral of csc squared three π₯ evaluated with respect to π₯.

To evaluate this integral, weβre going to need to recall a few trigonometric identities. First, we recall that csc π₯ is equal to one over sin π₯. And this means we can rewrite our integral as one over sin squared of three π₯ with respect to π₯. Next, weβre going to divide both the numerator and the denominator of this fraction by cos squared of three π₯. This might seem like a strange step. But we will see how it helps us in a moment. And the reason weβre allowed to perform this step is because multiplying or dividing both the numerator and the denominator by the same number gives us an equivalent fraction. It doesnβt change the size of the fraction.

So we now have the integral of one over cos squared of three π₯ over sin squared three π₯ over cos squared three π₯ with respect to π₯. But we know that one over cos of π₯ is equal to sec of π₯. And we also know that tan of π₯ is equal to sin of π₯ over cos of π₯. And we use this identity to rewrite one over cos squared of three π₯ as sec squared of three π₯. And we use this one to write sin squared three π₯ over cos squared three π₯ as tan squared of three π₯. And now, weβre ready to integrate.

We recall the fact that the derivative of tan of ππ₯, where π is a constant, is π times sec squared ππ₯. So weβre going to use integration by substitution. And we let π’ be equal to tan of three π₯ which means that dπ’ by dπ₯ is equal to three sec squared of three π₯. We need to rearrange this a little. We can say that the dπ’ is equal to three sec squared three π₯ dπ₯. And then, we divide by three. And we see that a third dπ’ is equal to sec squared of three π₯ dπ₯. Weβre now going to use this to replace the various parts of our integral. Letβs clear some space.

We replace one over tan squared three π₯ with one over π’ squared. Then, we replace sec squared three π₯ dπ₯ with a third dπ’. And we can now evaluate the integral of one over π’ squared times a third with respect to π’. Remember weβre allowed to take this constant of one-third outside of the integral and focus on integrating the expression in π’. And in fact, it can be useful to write one over π’ squared as π’ to the power of negative two. To integrate π’ to the negative two, we add one to the power and then we divide by that new number. So the integral of π’ to the negative two with respect to π’ is π’ to the power of negative one divided by negative one. And then, we have plus π. Thatβs our constant of integration. We let π’ be equal to tan of three π₯. So we can now replace π’ with tan of three π₯. And we get a third times tan of three π₯ to the negative one over negative one plus π.

Letβs lead this up a little bit and distribute our parentheses. We get negative one over three tan of three π₯ plus πΆ. Notice how I have actually replaced lowercase π with an uppercase πΆ as the constant itself has been multiplied by a third. So itβs now a different number. Finally, we recall the fact that one over tan of π₯ is equal to cot of π₯. And we can therefore write negative one over three tan of three π₯ as negative a third times cot of three π₯. And weβve seen that the integral of csc squared of three π₯ with respect to π₯ is negative one-third cot of three π₯ plus πΆ.