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int(cos2x-cos2alpha)/(cosx-cosalpha)dx...

`int(cos2x-cos2alpha)/(cosx-cosalpha)dx`

A

`sinx+xcosalpha+C`

B

`2sinx+xcosalpha+C`

C

`2sinx+2xcosalpha+C`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
C
`I=int((2cos^(2)x-1)-(2cos^(2)alpha-1))/((cosx-cosalpha))dx=2int((cos^(2)x-cos^(2)alpha))/((cosx-cosalpha))dx`.
`=2int(cosx+cosalpha)dx=int2cosxdx+2cosalpha*intdx`
`=sinx+2xcosalpha+C`.
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