Question Number 170871 by sciencestudent last updated on 02/Jun/22
$${Why}\:{is}\:{it}\:{equal}? \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\pi} {\int}}{sin}^{\mathrm{2}{p}} {udu}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{sin}^{\mathrm{2}{p}} {udu} \\ $$
Answered by thfchristopher last updated on 02/Jun/22
$$\int_{\mathrm{0}} ^{\pi} \mathrm{sin}^{\mathrm{2}{p}} {udu} \\ $$$$\mathrm{Let}\:{u}=\frac{\pi}{\mathrm{2}}−{x},\:{du}=−{dx} \\ $$$$\mathrm{When}\:{u}=\pi,\:{x}=−\frac{\pi}{\mathrm{2}} \\ $$$${u}=\mathrm{0},\:{x}=\frac{\pi}{\mathrm{2}} \\ $$$$\therefore\:\int_{\mathrm{0}} ^{\pi} \mathrm{sin}^{\mathrm{2}{p}} {udu} \\ $$$$=−\int_{\frac{\pi}{\mathrm{2}}} ^{-\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{2}{p}} \left(\frac{\pi}{\mathrm{2}}−{x}\right){dx} \\ $$$$=\int_{-\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2}{p}} {xdx} \\ $$$$\mathrm{Since}\:\mathrm{cos}^{\mathrm{2p}} \left(\frac{\pi}{\mathrm{2}}\right)=\mathrm{cos}^{\mathrm{2p}} \left(-\frac{\pi}{\mathrm{2}}\right), \\ $$$$\therefore\:\:\mathrm{it}\:\mathrm{is}\:\mathrm{an}\:\mathrm{even}\:\mathrm{function} \\ $$$$\Rightarrow\int_{-\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2}{p}} {xdx}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2}{p}} {xdx} \\ $$$$=−\mathrm{2}\int_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} \mathrm{cos}^{\mathrm{2}{p}} \left(\frac{\pi}{\mathrm{2}}−{u}\right){du} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{2}{p}} {udu} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\pi} \mathrm{sin}^{\mathrm{2}{p}} {udu}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{2}{p}} {udu} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \mathrm{sin}^{\mathrm{2}{p}} {udu}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{2}{p}} {udu} \\ $$