Evaluate the following integrals: ^ 2 _0 ^ -1 ( )dx - Vidyadip

Question

Evaluate the following integrals:$\int\limits^{2\pi}_0\cos^{-1}(\cos\text{x})\text{dx}$

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Answer

$\int\limits^{2\pi}_0\cos^{-1}(\cos\text{x})\text{dx}$$=\int\limits^{\pi}_0\cos^{-1}(\cos\text{x})\text{dx}+\int\limits^{2\pi}_\pi\cos^{-1}(\cos\text{x})\text{dx}$
$=\int\limits^{\pi}_0\text{x}\text{ dx}+\int\limits^{2\pi}_\pi(2\pi-\text{x})\text{dx}$ $\big[{\pi}\leq\text{x}\leq2\pi\Rightarrow-2\pi\leq-\text{x}\leq-{\pi}\Rightarrow0\leq2\pi-\text{x}\leq{\pi}\big]$
$=\Big[\frac{\text{x}^2}{2}\Big]+\bigg[\frac{(2\pi-\text{x})}{2\times(-1)}\bigg]^{2\pi}_\pi$
$=\frac{1}{2}(\pi^2-0)-\frac{1}{2}(0-\pi^2)$
$=\frac{\pi^2}{2}+\frac{\pi^2}{2}$
$=\pi^2$

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