### Video Transcript

Determine the indefinite integral
of negative three tan squared eight π₯ times csc squared eight π₯ evaluated with
respect to π₯. These does at first look quite
tricky. However, If we recall some of our
trigonometry identities, it does get a little nicer. We know that tan π₯ is equal to sin
π₯ over cos of π₯. And we also know that csc π₯ is
equal to one over sin π₯. We can, therefore, rewrite our
entire integrant as negative three times sin squared eight π₯ over cos squared eight
π₯ times one over sin squared eight π₯. And then we noticed that the sin
squared eight π₯ cancels. We can take the factor of negative
three outside of the integral sin to make the next step easier. And we have a negative three times
the integral of one over cos squared eight π₯ dπ₯.

But we know that one over cos of π₯
is equal to sec of π₯. So our integral becomes negative
three times the integral of sec squared eight π₯ evaluated with respect to π₯. But of course, the integral of sec
squared ππ₯, evaluated with respect to π₯, is one over π tan of ππ₯ plus some
constant of integration π. And so we see that the integral of
sec squared eight π₯ is an eighth tan of eight π₯ plus π. We distribute our parentheses. And we see that weβre left with
negative three-eights of tan of eight π₯ plus a new constant since we multiplied our
original one by negative three. Letβs call that capital πΆ.