Question

integrate (cos 2x - cos 2alpha)/(cos x - cos alpha)

Answer 1

Answer:

$2(sinx\ +\ xcos\alpha )\ +\ c$

Step-by-step explanation:

Given equation is

$\int {\frac{cos2x\ -\ cos2\alpha}{cosx\ -\ cos\alpha } } \, dx$

We know that

$cos2x = 2cos^{2}x - 1$

$cos2\alpha = 2cos^{2}\alpha\ -\ 1$

Put these value in above equation,

$= \int{\frac{2cos^{2}x\ -\ 1\ - (2cos^{2}\alpha\ -\ 1)}{cosx\ -\ cos\alpha  } \, dx$

$= \int{\frac{2cos^{2}x\ -\ 1\ -\ 2cos^{2}\alpha\ +\ 1}{cosx\ -\ cos\alpha  } \, dx$

$= 2\int{\frac{cos^{2}x\ -\ cos^{2}\alpha}{cosx\ -\ cos\alpha } } \, dx$

$= 2\int{\frac{(cosx\ +\ cos\alpha)(cosx\ -\ cos\alpha)}{(cosx\ -\ cos\alpha)} } \, dx$             ∵ $a^{2}\ -\ b^{2} = (a\ +\ b)(a\ -\ b)$

$= 2\int{(cosx\ +\ cos\alpha)} \, dx$

Integrating,

$= 2(sinx\ +\ xcos\alpha )\ +\ c$ [where c is a constant](Answer)

Note :- $where\ \alpha\ is\ a\ constant$