Question

Evaluate the integral `intcos^2xtan^3xdx`?

Answer 1

`intcos^2xtan^3xdx = 1/2cosx^2-ln|cosx| + C`

Explanation:

`intcos^2xtan^3xdx = intcos^2x(sin^3x/cos^3x)dx`
`:. intcos^2xtan^3xdx = intsin^3x/cosxdx`

We can perform this integral by a simple try substitution:

Let `u=cosx => (du)/dx=-sinx`, so `-int ... du=int...sinxdx`

`:. intcos^2xtan^3xdx = intsin^2xsinx/cosxdx`
`:. intcos^2xtan^3xdx = int(1-cos^2x)sinx/cosxdx`

Then if we perform the substitution we get:
`intcos^2xtan^3xdx = int(1-u^2)(-1/u)du`
`:. intcos^2xtan^3xdx = int(u^2-1)/udu`
`:. intcos^2xtan^3xdx = intu-1/udu`
`:. intcos^2xtan^3xdx = 1/2u^2-ln|u| + C`
`:. intcos^2xtan^3xdx = 1/2cosx^2-ln|cosx| + C`