`(1)/(1+cos alpha.cos x)`

Explanation

Let I `=int (1)/(1+cos alpha.cos x)dx`
`"Put tan "((x)/(2))=t therefore x=2tan^(-1)t`
`therefore dx=(2dt)/(1+t^(2)) and cos x=(1-t^(2))/(1+t^(2))`
`therefore I=int (1)/(1+(cos alpha)((1-t^(2))/(1+t^(2)))).(2dt)/(1+t^(2))`
`=int (1+t^(2))/(1+t^(2)+cos alpha-cos alpha.t^(2)).(2dt)/(1+t^(2))`
`=2int (1)/((1+cos alpha)+(1-cos alpha)t^(2))dt `
`=2int (1)/((2cos^(2)((alpha)/(2))+2sin^(2)((alpha)/(2)).t^(2)))dt`
`=(2)/(2sin^(2)((alpha)/(2)))int (1)/(cot^(2)((alpha)/(2))+t^(2))dt`
`=(1)/(sin^(2)((alpha)/(2)))xx(1)/(cot((alpha)/(2)))tan^(-1)[(t)/(cot((alpha)/(2)))]+c`
`=(2)/(sin^(2)((alpha)/(2)).cos((alpha)/(2))).tan^(-1)[tan((alpha)/(2)).tan((x)/(2))]+c`
`=2"cosec"alpha.tan^(-1)[tan((alpha)/(2)).tan((x)/(2))]+c`