Question
Evaluate the following integrals:$\int\limits^{{\pi}}_{-\frac{\pi}{2}}\sin^{-1}(\sin\text{x})\text{dx}$
✓
Answer
$\int\limits^{{\pi}}_{-\frac{\pi}{2}}\sin^{-1}\text{x}(\sin\text{x})\text{dx}$$=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin^{-1}(\sin\text{x})\text{dx}\int\limits^{{\pi}}_{\frac{\pi}{2}}\sin^{-1}(\sin\text{x})\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{x dx}+\int\limits^{{\pi}}_{\frac{\pi}{2}}(\pi-\text{x})\text{dx}$ $\Big[\frac{\pi}{2}\leq\text{x}\leq\pi\Rightarrow-\pi\leq-\text{x}\leq-\frac{\pi}{2}\Rightarrow0\leq\pi-\text{x}\leq\frac{\pi}{2}\Big]$
$=\Big[\frac{\text{x}^2}{2}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}+\bigg[\frac{(\pi-\text{x})}{2\times(-1)}\bigg]^{\pi}_{\frac{\pi}{2}}$
$=\frac{1}{2}\Big(\frac{\pi^2}{4}-\frac{\pi^2}{4}\Big)-\frac{1}{2}\Big(0-\frac{\pi^2}{4}\Big)$
$=0+\frac{\pi^2}{8}$
$=\frac{\pi^2}{8}$
$=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{x dx}+\int\limits^{{\pi}}_{\frac{\pi}{2}}(\pi-\text{x})\text{dx}$ $\Big[\frac{\pi}{2}\leq\text{x}\leq\pi\Rightarrow-\pi\leq-\text{x}\leq-\frac{\pi}{2}\Rightarrow0\leq\pi-\text{x}\leq\frac{\pi}{2}\Big]$
$=\Big[\frac{\text{x}^2}{2}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}+\bigg[\frac{(\pi-\text{x})}{2\times(-1)}\bigg]^{\pi}_{\frac{\pi}{2}}$
$=\frac{1}{2}\Big(\frac{\pi^2}{4}-\frac{\pi^2}{4}\Big)-\frac{1}{2}\Big(0-\frac{\pi^2}{4}\Big)$
$=0+\frac{\pi^2}{8}$
$=\frac{\pi^2}{8}$
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