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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 $-1$ :

$e^1-e^-1$