00:01
Hi, so today we're going to be solving the indefinite trig integral.
00:05
The integral we have is the sine of 2x, cost 3x, dx.
00:10
And the first thing we have to do is we're going to use a trig equation in order to manipulate this into something we can actually solve.
00:18
So what we're going to have is the equation that states the sign of a post -c is equal to one -half.
00:33
Times the sign of a plus b plus the sign of a minus b and so our 2x is with our sign so that's going to be our a and our 3x is with our cosines it's going to be our b and now we can just start substituting into the equation.
01:11
So we can take our one half out right away.
01:14
It's just a constant.
01:17
So we're going to have one half times the integral of the sign of 2x plus 3x plus the sign of 2x minus 3x.
01:41
And we have our dx of course.
01:47
And now from here we're just going to simplify.
01:49
So we're going to have one half times the integral of the sign of 5x.
02:00
And we can say plus the sign of minus x, dx.
02:09
And from here, we're going to just split this into two integrals.
02:11
So we're going to have one half time all of this.
02:16
So the integral of the sign of 5x.
02:21
And the sign of negative x is just equal to negative sign x.
02:28
So we're going to have minus the integral of the sign of x.
02:36
Sorry, i forgot our dx here.
02:41
Dx.
02:43
And now from here, all we have to do is we're going to do a u substitution for our first integral, and then we can solve both of our integral...