Verified answer

this is a really clever thing. it gets very messy depending on what exactly you are doing.

rewrite as

∫(1)cos^nxdx

now depending on what you are doing, you have two parts you can use :)

u=cos^nx

du=ncos^(n-1)xdx

dv=1dx

du=x

∫udv = uv - ∫vdu

∫(1)cos^nxdx = xcos^nx -∫xncos^(n-1)dx

the best way to learn this is to use integration by parts for

∫f '(x)dx

u=f'(x)

du=f"(x)dx

dv=1dx

v=x

∫f'(x)=xf'(x)-∫xf"(x)dx etc

and then do the same thing with this

∫f^n(x)

u=f^n(x)

du=f^(n-1)(x)

dv=1dx

v=x

∫f^n(x) = xf^n(x) -∫xf^(n-1)(x)dx

∫f^n(x) = xf^n)x) - [x^2f^(n-1) - ∫x^2(n-1)f^(n-1)(x)dx]

?x³ exp(x) dx = ?x³ d[exp(x)] = x³ exp(x) - ?exp(x) d[x³] = x³ exp(x) - ?exp(x) 3x² dx = x³ exp(x) - ? 3x² d[exp(x)] = x³ exp(x) - (3x² exp(x) - ? 3 exp(x) d[x²] ) = x³ exp(x) - (3x² exp(x) - ? 3 exp(x) 2x dx ) = end your self already