### Video Transcript

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

To evaluate this integral, weβre 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, weβre 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 weβre 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 doesnβt 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, weβre ready to integrate.

We recall the fact that the derivative of tan of ππ₯, where π is a constant, is π
times sec squared ππ₯. So weβre 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₯. Weβre now going to use this to replace the various parts of our integral. Letβs 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 weβre 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. Thatβs 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.

Letβs lead this up a little bit and distribute our parentheses. We get negative one over three tan of three π₯ plus πΆ. Notice how I have actually replaced lowercase π with an uppercase πΆ as the constant
itself has been multiplied by a third. So itβs 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 weβve seen that the integral of csc squared of three π₯ with respect to π₯ is
negative one-third cot of three π₯ plus πΆ.