Question
Evaluate the following integrals:
$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\text{x}\cos^2\text{x}\text{ dx}$
$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\text{x}\cos^2\text{x}\text{ dx}$
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Answer
We have,
$\text{I}=\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\text{x}\cos^2\text{x}\text{ dx}$
Let $\text{f}(\text{x})=\text{x}\cos^2\text{x}$
$\Rightarrow \text{f}(-\text{x})= (-\text{x})\cos^2(-\text{x})$
$= - \text{x}\cos^2\text{x}$
$\therefore \text{f}(-\text{x})=-\text{f}(\text{x})$
i.e., f(x) is odd function.
We know that $\int\limits^\text{a}_{-\text{a}}\text{f}(\text{x})\text{dx} = 0, $ if f(x) is odd function.
$\therefore\ \text{I}=\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\text{x}\cos^2\text{x}\text{ dx}=0$
$\text{I}=\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\text{x}\cos^2\text{x}\text{ dx}$
Let $\text{f}(\text{x})=\text{x}\cos^2\text{x}$
$\Rightarrow \text{f}(-\text{x})= (-\text{x})\cos^2(-\text{x})$
$= - \text{x}\cos^2\text{x}$
$\therefore \text{f}(-\text{x})=-\text{f}(\text{x})$
i.e., f(x) is odd function.
We know that $\int\limits^\text{a}_{-\text{a}}\text{f}(\text{x})\text{dx} = 0, $ if f(x) is odd function.
$\therefore\ \text{I}=\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\text{x}\cos^2\text{x}\text{ dx}=0$
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