Question

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

Answer 1

`-lnabs(1+cosx)+C`

Explanation:

We want to find

`intsinx/(1+cosx)dx`

To do so, we will use substitution: let `u=1+cosx`. This implies that `du=-sinxdx`.

To attain the `du=-sinxdx` value in the integral, multiply the interior of the integral by `-1`. Balance this by multiplying the exterior by `-1` as well.

`=-int(-sinx)/(1+cosx)dx`

Substituting in our `u` and `du` values, this gives us

`=-int1/udu`

This is a common integral:

`=-lnabsu+C`

Using our value for `u`:

`=-lnabs(1+cosx)+C`