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

Question

Isabella Ackerman · Accepted Answer

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

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#

Samantha Adkison · Answer

undefined To find the antiderivative of $ e^{\sin(x)} \cos(x) $, you can use integration by parts. Let $ u = e^{\sin(x)} $ and $ dv = \cos(x)dx $. Then, $ du = e^{\sin(x)} \cos(x) dx $ and $ v = \sin(x) $.

Applying the integration by parts formula:

$$ \int e^{\sin(x)} \cos(x) dx = e^{\sin(x)} \sin(x) - \int \sin(x) \cdot e^{\sin(x)} \cos(x) dx $$

This results in:

$$ \int e^{\sin(x)} \cos(x) dx = e^{\sin(x)} \sin(x) - \int e^{\sin(x)} \cos(x) dx $$

Now, you can add $ \int e^{\sin(x)} \cos(x) dx $ to both sides and solve for it:

$$ 2 \int e^{\sin(x)} \cos(x) dx = e^{\sin(x)} \sin(x) $$

$$ \int e^{\sin(x)} \cos(x) dx = \frac{1}{2} e^{\sin(x)} \sin(x) + C $$

Where $ C $ is the constant of integration.