what's integration of (tanx)^5?
Let's define
I_{n} = Integral of tan^n(x)dx
You can express I_{n} in terms of
I_{n-2} as follows. We write the integrand as:
tan^n(x) = sin^n(x)/cos^n(x)
Take out a factor sin^2(x):
tan^n(x) = sin^2(x)sin^(n-2)/cos^n(x)
And then substitute:
sin^2(x) = 1 - cos^2(x)
tan^n(x) = sin^(n-2)/cos^n(x) -
sin^(n-2)/cos^(n-2)(x)
So, we have:
I_{n} = - I{n-2} +
Integral of sin^(n-2)/cos^n(x) dx
We can compute the integral in this expression as follows. The integrand can be written as:
sin^(n-2)/cos^n(x) =
1/cos^2(x) tan^(n-2)(x)
So the integral is:
Integral of 1/cos^2(x)tan^(n-2)(x)dx =
Integral tan^(n-2) dtan(x) =
1/(n-1) tan^(n-1)(x)
The formula is thus:
I_{n} = - I{n-2} + 1/(n-1) tan^(n-1)(x)
This is valid for n > 1. For n = 1, we have:
I_{1} = Integral of tan(x) dx =
- Log|cos(x)|
Using the above formula you get for
n = 5:
I_{5} = - I_{3} + 1/4 tan^(4)(x)
For n = 3 you get:
I_{3} = -I_{1} + 1/2 tan^2(x) =
Log|cos(x)| + 1/2 tan^2(x)
So you get the result:
I_{5} = -Log|cos(x)| - 1/2 tan^2(x)+ 1/4 tan^(4)(x)