1.

`int ((1+tanx)/(1-tanx))dx`

Answer» Correct Answer - `-log |cosx 0 sinx| + C`
`I = int((1+(sinx)/(cosx))/(1-(sinx)/(cosx)))dx = int((cosx+sinx))/((cosx-sinx)) dx`
`= -int(dt)/(t)`, where `(cosx -sinx) =t`
`= - log|t| +C =-log|cosx-sinx|+C`.


Discussion

No Comment Found