Question #2ce29

1 Answer
Ratnaker Mehta
Dec 25, 2017

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.

Related questions
See all questions in Integrals of Exponential Functions
Impact of this question
1504 views around the world
You can reuse this answer
Creative Commons License