Question

What is the antiderivative of `(1+sinx)/(1-sinx)`?

Answer 1

`int(1+sin(x))/(1-sin(x))dx=2tan(x)+2/cos(x)-x+C`, `C in RR`

Explanation:

`I=int(1+sin(x))/(1-sin(x))dx`

`=int((1+sin(x))(1+sin(x)))/((1-sin(x))(1+sin(x)))dx`

`=int((1+sin(x))^2)/(1-sin(x)^2)dx`

Because `1-sin(x)^2=cos(x)^2`,

`I=int(sin(x)^2+2sin(x)+1)/(cos(x)^2)dx`

`=int(sin(x)^2)/(cos(x)^2)dx+2intsin(x)/(cos(x)^2)dx+int1/(cos(x)^2)dx`

`=inttan(x)^2dx+2intsin(x)/(cos(x)^2)dx+intsec(x)^2dx`

Now, let `u=cos(x)`

`du=-sin(x)dx`

`I=inttan(x)^2dx+intsec(x)^2dx-2int1/u^2du`

Also, `tan(x)^2=sec(x)^2-1`

`I=2intsec(x)^2dx-int1dx-2int1/u^2du`

`=2tan(x)+2/u-x`

`=2tan(x)+2/cos(x)-x+C`, `C in RR`

\0/ Here's our answer !