Question
Evaluate the following integrals:$\int^\limits{\frac{\pi}{2}}_{0}|\cos2\text{x}|\text{dx}$
✓
Answer
$\int^\limits{\frac{\pi}{2}}_{0}|\cos2\text{x}|\text{dx}$We know that,
$|\cos2\text{x}|=\begin{cases}-\cos2\text{x},&\frac{\pi}{4}\leq\text{x}\leq\frac{\pi}{2}\\\cos2\text{x},&0<\text{x}\leq\frac{\pi}{4}\end{cases}$
$\therefore\ \text{I}=\int^\limits{2}_{-2}|\cos2\text{x}|\text{dx}$
$\Rightarrow\text{I}=\int^\limits{\frac{\pi}{4}}_{0}\cos2\text{x }\text{dx}-\int^\limits{\frac{\pi}{2}}_{\frac{\pi}{4}}\cos2\text{x }\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{\sin2\text{x}}{2}\Big]^{\frac{\pi}{4}}_0-\Big[\frac{\sin2\text{x}}{2}\Big]^{\frac{\pi}{2}}_{\frac{\pi}{4}}$
$\Rightarrow\text{I}=\frac{1}{2}-0-0+\frac{1}{2}$
$\Rightarrow\text{I}=1$
$|\cos2\text{x}|=\begin{cases}-\cos2\text{x},&\frac{\pi}{4}\leq\text{x}\leq\frac{\pi}{2}\\\cos2\text{x},&0<\text{x}\leq\frac{\pi}{4}\end{cases}$
$\therefore\ \text{I}=\int^\limits{2}_{-2}|\cos2\text{x}|\text{dx}$
$\Rightarrow\text{I}=\int^\limits{\frac{\pi}{4}}_{0}\cos2\text{x }\text{dx}-\int^\limits{\frac{\pi}{2}}_{\frac{\pi}{4}}\cos2\text{x }\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{\sin2\text{x}}{2}\Big]^{\frac{\pi}{4}}_0-\Big[\frac{\sin2\text{x}}{2}\Big]^{\frac{\pi}{2}}_{\frac{\pi}{4}}$
$\Rightarrow\text{I}=\frac{1}{2}-0-0+\frac{1}{2}$
$\Rightarrow\text{I}=1$
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