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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

Whyisitequal?12π0sin2pudu=π20sin2pudu

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

0πsin2puduLetu=π2x,du=dxWhenu=π,x=π2u=0,x=π20πsin2pudu=π2π2sin2p(π2x)dx=π2π2cos2pxdxSincecos2p(π2)=cos2p(π2),itisanevenfunctionπ2π2cos2pxdx=20π2cos2pxdx=2π20cos2p(π2u)du=20π2sin2pudu0πsin2pudu=20π2sin2pudu120πsin2pudu=0π2sin2pudu

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