Question
Evaluate the following integrals:
$\int\limits^\pi_0\cos\text{x}|\cos\text{x}|\text{dx}$
$\int\limits^\pi_0\cos\text{x}|\cos\text{x}|\text{dx}$
✓
Answer
Consider $\text{f(x)}=\cos\text{x}|\cos\text{x}|$
Now,
$\text{f}(\pi-\text{x})=\cos(\pi-\text{x})|\cos(\pi-\text{x})|$
$=-\cos\text{x}|-\cos\text{x}|=-\cos\text{x}|\cos\text{x}|$
$=-\text{f(x)}$
$\therefore\ \int\limits^\pi_0\cos\text{x}|\cos\text{x}|\text{dx}=0$ $\begin{bmatrix}\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=\begin{cases}\int\limits^{2\text{a}}_0\text{f(x)}\text{dx},&\text{ if }\text{ f}(2\text{a}-\text{x})=\text{f(x)}\\0,&\text{ if }\text{ f}(2\text{a}-\text{x})=\text{f(x)}\end{cases}\end{bmatrix}$
Now,
$\text{f}(\pi-\text{x})=\cos(\pi-\text{x})|\cos(\pi-\text{x})|$
$=-\cos\text{x}|-\cos\text{x}|=-\cos\text{x}|\cos\text{x}|$
$=-\text{f(x)}$
$\therefore\ \int\limits^\pi_0\cos\text{x}|\cos\text{x}|\text{dx}=0$ $\begin{bmatrix}\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=\begin{cases}\int\limits^{2\text{a}}_0\text{f(x)}\text{dx},&\text{ if }\text{ f}(2\text{a}-\text{x})=\text{f(x)}\\0,&\text{ if }\text{ f}(2\text{a}-\text{x})=\text{f(x)}\end{cases}\end{bmatrix}$
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