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Answer 1 · 2017-04-15T19:30:21

#e^1-e^-1#

Use a #u# substitution.

Set #u=cosx#. #du=-sinxdx#

We don't have a negative so we'll put a negative on the outside of the integral and on the inside of the integral.

Making a #u# substitution will also cause a change in the integrand.

At #x=0#: #u=cos(0)=1#

At #x=pi#: #u=cos(pi)=-1#

So our problem is now:

#-int_1^-1 e^udu#

Use the negative on the outside to flip the integrand:

#int_-1^1 e^udu#

Take the integral. The integral of #e^u# will be #e^u#. Then plug in #1# and #-1#:

#e^1-e^-1#