Question

\int _ { \frac { \pi } { 6 } } ^ { \frac { \pi } { 3 } } 2 \sin 2 \theta d \theta​

Answer 1

Step-by-step explanation:

We have,

$\int _ { \frac { \pi } { 6 } } ^ { \frac { \pi } { 3 } } 2 \sin 2 \theta d \theta$

$=2 \int _ { \frac { \pi } { 6 } } ^ { \frac { \pi } { 3 } } \sin 2 \theta d \theta$

$= - 2 \bigg [\frac{ \cos 2 \theta }{2} \bigg ]_ { \frac { \pi } { 6 } } ^ { \frac { \pi } { 3 } }$

$= - \bigg [ \cos \frac{2\pi}{3} - \cos \frac{2\pi}{6} \bigg ]$

$= - \bigg [ - \frac{1}{2} - \frac{1}{2} \bigg ]$

$= 1$