Evaluate: `int1/(sinx(2cos^2x-1))\ dx`

Explanation

Putting `cosx=t`, we get
`I=int(1/(sinx(2cos^(2)x-1)))dx=int(1/(sinx(2t^(2)-1))xx -(dt)/(sinx))=-int1/((1-t^(2)(2t^(2)-1))`dt
`therefore I=-int(1/(1-t^(2))+2/(2t^(2)-1))dt=-int1/(1-t^(2))dt-2int1/(2t^(2)-1)dt`
`=-1/2ln |(1+t)/(1-t)|-sqrt(2)/2ln |(sqrt(2)t-1)/(sqrt(2)t+1)|+C=-1/2ln|(1+cosx)/(1-cosx)|-1/sqrt(2)ln|(sqrt(2)cosx-1)/(sqrt(2)cosx+1)|+C`