Question

Integral of sec (3x) tan (3x) dx

Answer 1

`\int {sec (3x)tan(3x)} \, dx= \int { (1)/(cos(3x))* (sin(3x))/(cos(3x)) } \, dx`Substitution:u=cos(3x), du=-3sin(3x)dx, sin(3x)dx= du/-3 Integral becomes:
`(-1)/(3) *\int { (1)/( u^(2) ) } \, du= (-1)/(3) \int {u^(-2) } \, du=( u^(-1) )/(3)= (1)/(3u)= (1)/(3cos(3x))+C`