Question

Integrate the following functions with respect to x :

`(cos2x - cos 2 alpha)/(cosx - cos alpha)`

Answer 1

`int((cos2x - cos2 alpha)/(cosx - cos alpha)) "d"x`

= `int [((2 cos^2x - 1) - (2cos^2alpha - 1))/(cosx - cosalpha)]* "d"x`

cos 2x = cos²x – sin²x – 1

cos 2x = 2 cos²x – 1

= `int (2cos^2x - 1 - 2cos^2alpha + 1)/(cosx - cosalpha) * "d"x`

= `int (2cos^x - 2cos^2alpha)/(cosx - cosalpha) * "d"x`

= `int (2(cos^2x - cos^2alpha))/(cosx - cosalpha) * "d"x`

= `2 int [((cosx + cosalpha)(cosx - cosalpha))/(cosx - cosalpha)] * "d"x`

= `2 int(cosx + cosalpha) "d"x`

= `2 int cos x  "d"x + 2 int cos alpha  "d"x`

= `2 int cos x  "d"x + 2 int cos alpha int  "d"x`

= 2 sin x + 2 cos α(x) + c

= 2 sin x + 2x cos α + c