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Question Number 42810 by maxmathsup by imad last updated on 02/Sep/18
find∫0∞x51+x7dx.
Commented by prof Abdo imad last updated on 03/Sep/18
changementx=t17giveI=∫0∞t571+t17t17−1dt=17∫0∞t67−11+tdt=17πsin(6π7)=π7sin(π7).
Answered by tanmay.chaudhury50@gmail.com last updated on 03/Sep/18
x7=tan2αx=(tanα)27dx=27tan−57αsec2αdα∫0Π2(tanα)107sec2α×27(tanα)−57×sec2αdα27∫0Π2(tanα)57αdα27∫0Π2(sinα)57(cosα)−57dαusingformula∫0Π2sin2p−1αcos2q−1αdα=⌈p×⌈q2⌈(p+q)here2p−1=572p=127p=672q−1=−572q=27q=17=27∫0Π2(sinα)2×67−1(cosα)2×17−1dα27×⌈(67)×⌈(17)2⌈(67+17)=17×⌈(17)×⌈(1−17)=17×Πsin(Π7)⌈(p)×⌈(1−p)=Πsin(pΠ)
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