Answer
$=\frac{2+\sqrt 2}{2}$
Work Step by Step
$\int^\frac{3\pi}{4}_0\sin(\theta) d\theta$
Since we know that the integral of $sin(\theta)$ is $-cos(\theta)$, we require:
$[-cos(\theta)]^\frac{3\pi}{4}_0$
This is equal to:
$-cos(\frac{3\pi}{4})-(-cos(0))$
Re-arranging and simplifying, we get:
$cos(0)-cos(\frac{3\pi}{4})$
Using exact values and radian measure, this can be simplified to:
$1-(-\frac{1}{\sqrt 2})$
$=1+\frac{1}{\sqrt 2}$
By simplifying then rationalising the denominator:
$\frac{\sqrt 2 +1}{\sqrt 2}$
$=\frac{\sqrt 2 +1}{\sqrt 2} \times \frac{\sqrt 2}{\sqrt 2}$
$=\frac{\sqrt 2(\sqrt 2 +1)}{2}$
$=\frac{2+\sqrt 2}{2}$
