Question

Evaulate:
integral 3x (sinx/cos^4x) dx


I think it's sec³ x , but that from using a piece of software, so you'll have to verify that.
Using uppercase 's' for the integral sign we have
S 3sin(x)/cos⁴dx or
S cos^-4(x)*3sin(x)dx
If you let u = cos(x) then du = -sin(x)dx
So the integral becomes
-3 S u^-4du = -3 * (1/-3)u^-3 = 1/cos³ x
sec³ x