How do you find the definite integral for: #e^sin(x) * cos(x) dx# for the intervals #[0, pi/4]#?

1 Answer
Jun 27, 2016

Use a #u#-substitution to get #int_0^(pi/4) e^sinx*cosxdx=e^(sqrt(2)/2)-1#.

Explanation:

We'll begin by solving the indefinite integral and then deal with the bounds.

In #inte^sinx*cosxdx#, we have #sinx# and its derivative, #cosx#. Therefore we can use a #u#-substitution.

Let #u=sinx->(du)/dx=cosx->du=cosxdx#. Making the substitution, we have:
#inte^udu#
#=e^u#

Finally, back substitute #u=sinx# to get the final result:
#e^sinx#

Now we can evaluate this from #0# to #pi/4#:
#[e^sinx]_0^(pi/4)#
#=(e^sin(pi/4)-e^0)#
#=e^(sqrt(2)/2)-1#
#~~1.028#