Evaluate the definite integral.?

Apr 24, 2018

$0$

Explanation:

This can be solved using the following substitution:

$u = {x}^{2}$

Since this is a definite, calculate the new lower and upper bounds using the substitution:

Lower: $u = {0}^{2} = 0$

Upper: $u = {\left(\sqrt{\pi}\right)}^{2} = \pi$

Take the differential:

$\mathrm{du} = 2 x \mathrm{dx}$

We see that $x \mathrm{dx}$ is present in the integral. As a result, we can divide both sides of the differential by $2$:

$\frac{1}{2} \mathrm{du} = x \mathrm{dx}$

And we get

$\frac{1}{2} {\int}_{0}^{\pi} \cos u \mathrm{du} = \frac{1}{2} \left(\sin u {|}_{0}^{\pi}\right)$

$= \frac{1}{2} \left(\sin \pi - \sin 0\right) = \frac{1}{2} \left(0 - 0\right) = 0$