Question #2ce29

1 Answer
Dec 25, 2017

Answer:

# 2sinx+2xcosalpha+C#.

Explanation:

Let, #I=int(cos2x-cos2alpha)/(cosx-cosalpha)dx#.

Then, #I=int{(2cos^2x-1)-(2cos^2alpha-1)}/(cosx-cosalpha)dx#,

#=2int(cos^2x-cos^2alpha)/(cosx-cosalpha)dx#,

#=2int{(cosx-cosalpha)(cosx+cosalpha)}/(cosx-cosalpha)dx#,

#=2int(cosx+cosalpha)dx=2intcosxdx+2cosalphaint1dx#.

# rArr I=2sinx+2xcosalpha+C#.