00:01
All right, so we need to calculate the antide derivative of this function here, tangent to the power of 5 of x over 2.
00:08
In order to do so, we're going to rewrite this integral in terms of integrals we already know how to calculate.
00:16
So the first integral that we're going to need to know is this one up here.
00:21
So when we have a function of the form tangent to some power n times secant square of x, this is the derivative of tangent.
00:31
Know that antiderivative is of this form.
00:35
Now, the second antedributive we're going to need to know is that of tangent, and that's given to us right here.
00:43
So what we're going to do first is we're going to split this integrand into two factors.
00:54
So we can write integral of tangent cubed of x over 2 times the tangent squared of x over 2 d x.
01:10
Now we can use this trigometric identity here to substitute in secant squared of x minus 1 for tangent squared of x that's going to give us integral tangent cubed x over 2 times secant squared of x minus 1 d x.
01:39
Now, we can multiply this factor here into this term, and then split it into two integrals.
01:50
Let's go ahead and do that.
01:52
So we have first integral is going to be tangent cubed x over 2, secant squared of x over 2, and then we have minus the integral of tangent cubed, x over 2, and then we have minus the integral of tangent cubed...