00:01
This problem is from chapter 7 and section 2, problem number 31 from the book calculus, early transcendental's 8th edition by james stewart.
00:11
And we have an indefinite integral of tangent to the fifth power of x.
00:17
One way to proceed here is to break this tangent to the fifth power by writing it as tangent cubed times tangent squared.
00:35
And for the tangent squared, we can use the pythagran identity over here.
00:41
On the right to rewrite this as secan squared x minus 1.
00:53
So doing so gives us tan cubed x times secant squared x minus tan cubed x minus tan cubed times 1.
01:15
It looks like for this first integral here we can use a u substitution because we have a secan squared.
01:22
Not quite for the second integral, at least not yet.
01:24
So let's go ahead and break this into two separate integrals.
01:27
So here let's get color coordinated.
01:32
So let's do the first integral in blue.
01:37
Tn cubed, secan squared minus and we have an integral tangent cubed.
01:51
And let's go ahead and write that tangent cubed is tan squared times tangent.
02:04
And the reason for doing so is once again we can rewrite this tan squared as secan squared minus one.
02:16
For our first integral in blue, we see that we can use.
02:21
A u substitution, let's take u to be tangent, then du is secan squared dx.
02:44
So let's maybe pick up our inequality down here.
02:50
So after this u substitution, we have u cubed, and we still have the second integral here, which after using this pythagran identity, we have secan squared x minus one.
03:14
Times tangent.
03:24
For this first integral we can go ahead and just use the power rule and we get u to the fourth power over 4...