WEBVTT
00:00:00.352 --> 00:00:09.182
Determine the indefinite integral of negative three tan squared eight π₯ times csc squared eight π₯ evaluated with respect to π₯.
00:00:09.522 --> 00:00:12.002
These does at first look quite tricky.
00:00:12.202 --> 00:00:16.652
However, If we recall some of our trigonometry identities, it does get a little nicer.
00:00:17.082 --> 00:00:20.832
We know that tan π₯ is equal to sin π₯ over cos of π₯.
00:00:21.002 --> 00:00:24.992
And we also know that csc π₯ is equal to one over sin π₯.
00:00:25.362 --> 00:00:34.932
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₯.
00:00:35.082 --> 00:00:38.402
And then we noticed that the sin squared eight π₯ cancels.
00:00:38.702 --> 00:00:43.392
We can take the factor of negative three outside of the integral sin to make the next step easier.
00:00:43.572 --> 00:00:48.682
And we have a negative three times the integral of one over cos squared eight π₯ dπ₯.
00:00:48.952 --> 00:00:52.182
But we know that one over cos of π₯ is equal to sec of π₯.
00:00:52.462 --> 00:00:59.242
So our integral becomes negative three times the integral of sec squared eight π₯ evaluated with respect to π₯.
00:00:59.502 --> 00:01:08.262
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.
00:01:08.662 --> 00:01:15.502
And so we see that the integral of sec squared eight π₯ is an eighth tan of eight π₯ plus π.
00:01:16.062 --> 00:01:17.642
We distribute our parentheses.
00:01:17.842 --> 00:01:26.332
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.
00:01:26.592 --> 00:01:28.042
Letβs call that capital πΆ.