Answer 1

`"I" = int Sinx/(sqrt(cos^2 x-2 cos x-3)) dx`

Put cos x = t
∴ – sin x dx = dt
sin x dx = – dt 

 =`int (-dt)/sqrt( t^2-2  t-3)` 

=`-int dt/sqrt( t^2-2  t + 1- 1-3)`

=`-int dt /sqrt((t-1)^2-(2)^2`                ....(Completing the square)

=`-log|(t-1)+sqrt((t-1)^2-2^2)|+C`       ....`{int(dx)/sqrt(x^2-a^2)=log|x+sqrt(x^2-a^2)|+"C"}`

=`-log|t-1+sqrt(t^2-2t-3)|+C`

`=– log|(cosx-1)+sqrt(cos^2x-2cosx-3)|+C`