Integrate? Sin^2 3x + 4cos4x

1 Answer
Dec 29, 2017

#x/2-(sin6x)/12+sin4x+C#

Explanation:

.

#int(sin^2(3x)+4cos4x)dx=intsin^2(3x)dx+int4cos4xdx=#

#int(1-cos6x)/2dx+4intcos4xdx=#

#1/2int(1-cos6x)dx+4(1/4)sin4x=#

#1/2intdx-1/2intcos6xdx+sin4x=#

#1/2x-1/2(1/6)sin6x+sin4x=#

#x/2-(sin6x)/12+sin4x+C#