Question

Question #9e12d

Answer 1

`-lnabscos(-e^-x)+C`

Explanation:

I'll be using the common notation `"tg"(x)=tan(x)`. So, you're asking for

`inte^-xtan(-e^-x)dx`

Let `u=-e^-x`. Differentiating this shows that `du=e^-xdx`:

`=inttan(-e^-x)(e^-xdx)=inttan(u)du`

Rewriting the tangent function:

`=intsin(u)/cos(u)du`

Now let `v=cos(u)`. This implies that `dv=-sin(u)du`. Our integrand has both `v` and is is a factor of `-1` away from also containing `dv`:

`=-int(-sin(u))/cos(u)du=-int1/vdv`

This is a well-known and important antiderivative:

`=-lnabsv`

Returning to our original variable `x` using `v=cos(u)` and `u=-e^-x`:

`=-lnabscos(u)=-lnabscos(-e^-x)+C`