Question

Use the derivative (below) to find the integral of `cos^nx dx`?

The question asks to find `d/dx(sinx*cos^(n-1)x)` which is just `cos^nx +(1-n)*sin^2x*cos^(n-2)x`
and then to use that to show that `n int cos^nx dx = sinx*cos^(n-1)x + (n-1) int cos^(n-2)x dx` but I get a `sin^2x` in with the second integral there. What is the problem going on here? Thank you!

Answer 1

Using the product rule in fact:

`d/dx (sinx cos^(n-1)x) = cos^nx - (n-1)sin^2x cos^(n-2)x`

Integrating both sides of the equality and using the linearity of the integral:

`sinx cos^(n-1)x = int cos^nx dx - (n-1)int sin^2x cos^(n-2)x dx`

use now the trigonometric identity:

`sin^2x = 1-cos^2x`

to get:

`sinx cos^(n-1)x = int cos^nx dx - (n-1)int (1-cos^2x) cos^(n-2)x dx`

`sinx cos^(n-1)x = int cos^nx dx - (n-1)int cos^(n-2)x dx +int (n-1)cos^nx dx`

`sinx cos^(n-1)x = n int cos^nx dx - (n-1)int cos^(n-2)x dx`

and finally:

`n int cos^nx dx = sinx cos^(n-1)x + (n-1)int cos^(n-2)x dx`