Answer
$\displaystyle \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sin{2x}\ dx = 0$
Work Step by Step
To evaluate the integral $\displaystyle \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sin{2x}\ dx$ we use the trigonometric identity $\sin{2x} = 2\sin{x}\cos{x}$ and let $u = \sin{x}$.
Then we obtain $u = \sin{x} \longrightarrow du = \cos{x}\ dx$
Before substituting, determine the new upper and lower limits of integration
Lower limit: When $x = -\dfrac{\pi}{4}$, $u = \sin{\Big(-\dfrac{\pi}{4}\Big)} = -\dfrac{\sqrt{2}}{2}$
Upper limit: When $x = \dfrac{\pi}{4}$, $u = \sin{\Big(\dfrac{\pi}{4}\Big)} = \dfrac{\sqrt{2}}{2}$
$\displaystyle \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sin{2x}\ dx = \displaystyle \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}2\sin{x}\cos{x}\ dx = \displaystyle 2\int_{-\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}}u\ du = 2\dfrac{u^2}{2}\Bigg\vert_{-\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} = u^2\Bigg\vert_{-\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}}= \bigg(\dfrac{\sqrt{2}}{2}\bigg)^2 - \bigg(-\dfrac{\sqrt{2}}{2}\bigg)^2 = \dfrac{2}{4} - \dfrac{2}{4} = 0$
