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Question Number 170871 by sciencestudent last updated on 02/Jun/22

Why is it equal? (1/2)∫_0 ^π sin^(2p) udu=∫_0 ^(π/2) sin^(2p) udu

$${Why}\:{is}\:{it}\:{equal}? \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\pi} {\int}}{sin}^{\mathrm{2}{p}} {udu}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{sin}^{\mathrm{2}{p}} {udu} \\ $$

Answered by thfchristopher last updated on 02/Jun/22

∫_0 ^π sin^(2p) udu Let u=(π/2)−x, du=−dx When u=π, x=−(π/2) u=0, x=(π/2) ∴ ∫_0 ^π sin^(2p) udu =−∫_(π/2) ^(-(π/2)) sin^(2p) ((π/2)−x)dx =∫_(-(π/2)) ^(π/2) cos^(2p) xdx Since cos^(2p) ((π/2))=cos^(2p) (-(π/2)), ∴ it is an even function ⇒∫_(-(π/2)) ^(π/2) cos^(2p) xdx=2∫_0 ^(π/2) cos^(2p) xdx =−2∫_(π/2) ^0 cos^(2p) ((π/2)−u)du =2∫_0 ^(π/2) sin^(2p) udu ⇒∫_0 ^π sin^(2p) udu=2∫_0 ^(π/2) sin^(2p) udu ⇒(1/2)∫_0 ^π sin^(2p) udu=∫_0 ^(π/2) sin^(2p) udu

$$\int_{\mathrm{0}} ^{\pi} \mathrm{sin}^{\mathrm{2}{p}} {udu} \\ $$$$\mathrm{Let}\:{u}=\frac{\pi}{\mathrm{2}}−{x},\:{du}=−{dx} \\ $$$$\mathrm{When}\:{u}=\pi,\:{x}=−\frac{\pi}{\mathrm{2}} \\ $$$${u}=\mathrm{0},\:{x}=\frac{\pi}{\mathrm{2}} \\ $$$$\therefore\:\int_{\mathrm{0}} ^{\pi} \mathrm{sin}^{\mathrm{2}{p}} {udu} \\ $$$$=−\int_{\frac{\pi}{\mathrm{2}}} ^{-\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{2}{p}} \left(\frac{\pi}{\mathrm{2}}−{x}\right){dx} \\ $$$$=\int_{-\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2}{p}} {xdx} \\ $$$$\mathrm{Since}\:\mathrm{cos}^{\mathrm{2p}} \left(\frac{\pi}{\mathrm{2}}\right)=\mathrm{cos}^{\mathrm{2p}} \left(-\frac{\pi}{\mathrm{2}}\right), \\ $$$$\therefore\:\:\mathrm{it}\:\mathrm{is}\:\mathrm{an}\:\mathrm{even}\:\mathrm{function} \\ $$$$\Rightarrow\int_{-\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2}{p}} {xdx}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2}{p}} {xdx} \\ $$$$=−\mathrm{2}\int_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} \mathrm{cos}^{\mathrm{2}{p}} \left(\frac{\pi}{\mathrm{2}}−{u}\right){du} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{2}{p}} {udu} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\pi} \mathrm{sin}^{\mathrm{2}{p}} {udu}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{2}{p}} {udu} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \mathrm{sin}^{\mathrm{2}{p}} {udu}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{2}{p}} {udu} \\ $$

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