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Question Number 170871 by sciencestudent last updated on 02/Jun/22
Whyisitequal?12∫π0sin2pudu=∫π20sin2pudu
Answered by thfchristopher last updated on 02/Jun/22
∫0πsin2puduLetu=π2−x,du=−dxWhenu=π,x=−π2u=0,x=π2∴∫0πsin2pudu=−∫π2−π2sin2p(π2−x)dx=∫−π2π2cos2pxdxSincecos2p(π2)=cos2p(−π2),∴itisanevenfunction⇒∫−π2π2cos2pxdx=2∫0π2cos2pxdx=−2∫π20cos2p(π2−u)du=2∫0π2sin2pudu⇒∫0πsin2pudu=2∫0π2sin2pudu⇒12∫0πsin2pudu=∫0π2sin2pudu
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