Question

How do you integrate `sin(x)tan(x)`?

Answer 1

`(sin(x))^3/3+C`

Explanation:

Writing your integral in the form `\int sin(x)^2/cos(x)dx` and substitute`t=sin(x)`
then we get
`dx=dt/cos(x)`

Answer 2

The answer is `=ln(|secx+tanx|)-sinx+C`

Explanation:

The integral is

`I=intsinxtanxdx=int(sin^2xdx)/cosx`

`=int(1-cos^2x)secxdx`

`=int(secx-cosx)dx`

`=intsecxdx-intcosxdx`

`=I_1+I_2`

First calculate

`I_1=intsecxdx`

`=int(secx(secx+tanx)dx)/(secx+tanx)`

`=int((sec^2x+secxtanx)dx)/(secx+tanx)`

Let `u=secx+tanx`, `=>`, `du=(secxtanx+sec^2x)dx`

Therefore,

`I_1=int(du)/u=lnu`

`=ln(secx+tanx)`

Second calculate

`I_2=intcosxdx=sinx`

And finally,

`I=ln(|secx+tanx|)-sinx+C`