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Question Number 42810 by maxmathsup by imad last updated on 02/Sep/18

find  ∫_0 ^∞     (x^5 /(1+x^7 ))dx  .

find0x51+x7dx.

Commented by prof Abdo imad last updated on 03/Sep/18

changement  x=t^(1/7)    give   I  = ∫_0 ^∞    (t^(5/7) /(1+t)) (1/7) t^((1/7)−1) dt  =(1/7) ∫_0 ^∞     (t^((6/7)−1) /(1+t)) dt =(1/7) (π/(sin(((6π)/7)))) = (π/(7sin((π/7)))) .

changementx=t17giveI=0t571+t17t171dt=170t6711+tdt=17πsin(6π7)=π7sin(π7).

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Sep/18

x^7 =tan^2 α   x=(tanα)^(2/7)   dx=(2/7)tan^((−5)/7) α sec^2 α dα  ∫_0 ^(Π/2) (((tanα)^((10)/7) )/(sec^2 α))×(2/7)(tanα)^((−5)/7) ×sec^2 αdα  (2/7)∫_0 ^(Π/2) (tanα)^(5/7) αdα  (2/7)∫_0 ^(Π/2) (sinα)^(5/7) (cosα)^((−5)/7) dα  using formula  ∫_0 ^(Π/2) sin^(2p−1) αcos^(2q−1) α dα =((⌈p ×⌈q)/(2⌈(p+q)))  here 2p−1=(5/7)  2p=((12)/7)   p=(6/7)  2q−1=−(5/7)   2q=(2/7)   q=(1/7)  =(2/7)∫_0 ^(Π/2) (sinα^ )^(2×(6/7)−1) (cosα)^(((2×1)/7)−1) dα  (2/7)×((⌈((6/7))×⌈((1/7)))/(2⌈((6/7)+(1/7))))=(1/7)×⌈((1/7))×⌈(1−(1/7))=(1/7)×(Π/(sin((Π/7))))    ⌈(p)×⌈(1−p)=(Π/(sin(pΠ)))

x7=tan2αx=(tanα)27dx=27tan57αsec2αdα0Π2(tanα)107sec2α×27(tanα)57×sec2αdα270Π2(tanα)57αdα270Π2(sinα)57(cosα)57dαusingformula0Π2sin2p1αcos2q1αdα=p×q2(p+q)here2p1=572p=127p=672q1=572q=27q=17=270Π2(sinα)2×671(cosα)2×171dα27×(67)×(17)2(67+17)=17×(17)×(117)=17×Πsin(Π7)(p)×(1p)=Πsin(pΠ)

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