Trigonometry and Integrals

Question

Let n be a positive integer at least 3 in value, and let I_n=∫1/cos^n(x) dx. Use integration by parts to obtain that I_n = (n-2/n-1) × I_n-2 + 1/n-1 + sin(x)cos^I-n(x)

Answer 1

Here is my solution, which differs from what you are supposed to prove in that I got multiplication, where the given result has addition.

I believe that it is a typo on their part, but you should be the judge.

I_n = ∫1/cosⁿ(x)dx = ∫cos^-2(x)cos^2-n(x)dx

Use integration by parts following the pattern

∫u'v = uv - ∫uv'

where

u' = cos^-2(x)

v = cos^2-n(x)

Thus

u = tan(x)

v' = (2-n)cos^1-n(x)(-sin(x))

I_n = ∫cos^-2(x)cos^2-n(x)dx =

tan(x)cos^2-n(x) + ∫tan(x)(2-n)cos^1-n(x)sin(x)dx =

sin(x)cos^1-n(x) + (2-n)∫sin²(x)cos^-n(x)dx =

sin(x)cos^1-n(x) + (2-n)∫(1-cos²(x))cos^-n(x)dx =

sin(x)cos^1-n(x) + (2-n)∫cos^-n(x)dx - (2-n)∫cos^2-n(x)dx =

sin(x)cos^1-n(x) + (2-n)I_n - (2-n)I_n-2

Collect I_n on the left of the equation

(n-1)I_n = (n-2)I_n-2 + sin(x)cos^1-n(x)

I_n = ((n-2)/(n-1))I_n-2 + (1/(n-1))sin(x)cos^1-n(x)