Question

How do you find the antiderivative of `e^(sinx)*cosx`?

Answer 1

Use a `u`-substitution to find `inte^sinx*cosxdx=e^sinx+C`.

Explanation:

Notice that the derivative of `sinx` is `cosx`, and since these appear in the same integral, this problem is solved with a `u`-substitution.

Let `u=sinx->(du)/(dx)=cosx->du=cosxdx`

`inte^sinx*cosxdx` becomes:
`inte^udu`

This integral evaluates to `e^u+C` (because the derivative of `e^u` is `e^u`). But `u=sinx`, so:
`inte^sinx*cosxdx=inte^udu=e^u+C=e^sinx+C`