Evaluate: `int(sinx)/(cos2x)\ dx`

Answer

` -(1)/(2sqrt(2))"log"_(e)|(sqrt(2)t-1)/(sqrt(2)t+1)|+C`

Explanation

`I=int(sinx)/(cos2x)dx=int(sinx)/(2cos^(2)x-1)dx`
` "Let " cosx=t`
` :. -sinx dx=dt`
` :. I=int(-dt)/(2t^(2)-1)= -(1)/(2)int(dt)/(t^(2)-(1)/(2))`
`=-(1)/(2)*(1)/(2(1)/(sqrt(2)))"log"_(e)|(t-(1)/(sqrt(2)))/(t+(1)/(sqrt(2)))|+C`
`= -(1)/(2sqrt(2))"log"_(e)|(sqrt(2)t-1)/(sqrt(2)t+1)|+C`