How do you evaluate the definite integral #int e^(sin(x))*cosx dx# from #[0,pi]#?

1 Answer
Mar 26, 2016

#0#

Explanation:

We have

#int_0^pie^sinxcosxdx#

If we let #u=sinx#, then #du=cosxdx#.

We will want to change the bounds by plugging the current bounds into our #u=sinx# expression.

#u(0)=sin(0)=0#
#u(pi)=sin(pi)=0#

Notice that something fairly odd has happened--both our bounds have become #0#. We have the definite integral

#=int_0^0e^udu#

When both the bounds of an integral are the same, the integral's value is #0#.

Recall that the integral just finds the area under a curve--there is no area if we are traveling from #x=0# to #x=0#, since we haven't moved at all. This is essentially a rectangle with width #0#.