Question Number 100087 by mathmax by abdo last updated on 24/Jun/20
Commented bybemath last updated on 25/Jun/20
![∫(cos ^2 x−1)(2+cos x)^6 d(cos x) ∫(z^2 −1)(2+z)^6 dz ∫ z^2 (2+z)^6 dz−∫(2+z)^6 dz [(1/7)z^2 (2+z)^7 −(1/(28))z(2+z)^8 +(1/(494))(2+z)^9 −(1/7)(2+z)^7 ]_1 ^(−1) [ (1/7)+(1/(28))+(1/(494))−(1/7)]−[(3^7 /7)−(3^8 /(28))+(3^9 /(494))−(3^7 /7)] = ((3^8 +1)/(28))+((1−3^9 )/(494)) ■](Q100128.png)
Answered by 1549442205 last updated on 26/Jun/20
![F=−∫_0 ^π (1−cos^2 x)(2+cos x)^6 dcos x =_(t=cos x ) ∫(t^2 −1)(t+2)^6 dt=∫[t^2 (t+2)^6 ]d∫t−∫(t+2)^6 dt=A−B A=∫t^2 (t^6 +12t^5 +60t^4 +160t^3 +240t^2 +192t+64)dt =(t^9 /9)+((12t^8 )/8)+((60t^7 )/7)+((160t^6 )/6)+((240t^5 )/5)+((192t^4 )/4)+((64t^3 )/3) B=(t^7 /7)+((12t^6 )/6)+((60t^5 )/5)+((160t^4 )/4)+((240t^3 )/3)+((192t^2 )/2)+64t Hence,∫_0 ^π sin^3 x(2+cos x)^6 dx=(A−B)∣_1 ^(−1) =((t^9 /9)+((12t^8 )/8)+((59t^7 )/7)+((148t^6 )/6)+((180t^5 )/5)+((32t^4 )/4)−((176t^3 )/3)−((192t^2 )/2)−64t)∣_1 ^(−1) =((2053)/(126))−(−((17635)/(126)))=((19688)/(126))=((9844)/(63))≈156.25 The second way: F=∫_0 ^π sin^3 x(2+cos x)^6 dx=∫_0 ^π sin^3 x(cos^6 x+12cos^5 x+60cos^4 x+160cos^3 x+240cos^2 x+192cos x+64)dx Since,∫_0 ^π sin^(2m+1) xcos^(2n+1) xdx= =∫_0 ^(π/2) sin^(2m+1) xcos^(2n+1) xdx+∫_(π/2) ^π sin^(2m+1) x.cos^(2n+1) xdx =_(x=t+(π/2) ) ∫_0 ^(π/2) sin^(2m+1) x.cos^(2n+1) xdx−∫_0 ^(π/2) cos^(2m+1) t.sin^(2n+1) tdt (1/2)B(m+1,n+1)−(1/2)B(n+1,m+1)=0(due to the property of Beta function:B(p,q)=B(q,p) and ,∫_0 ^π sin^(2m) xcos^(2n+1) xdx= =∫_0 ^(π/2) sin^(2m) xcos^(2n+1) xdx+∫_(π/2) ^π sin^(2m) x.cos^(2n+1) xdx ==∫_0 ^(π/2) sin^(2m) xcos^(2n+1) xdx+∫_0 ^(π/2) cos^(2m) x.sin^(2n+1) xdx= (1/2)B(((2m+1)/2),n+1)+(1/2)B(n+1,((2m+1)/2))=B(((2m+1)/2),n+1),so F=∫_0 ^π sin^3 x(cos^6 x+12cos^5 x+60cos^4 x+160cos^3 x+240cos^2 x+192cos x+64)dx =B(2,(7/2))+60B(2,(5/2))+240B(2,(3/2))+64B(2,(1/2)) B(p,q)=((Γ(p).Γ(q))/(Γ(p+q))),so B(2,(7/2))=((Γ(2).Γ((7/2)))/(Γ(2+(7/2)))) Γ(2)=1,Γ((1/2))=(√π),Γ((3/2))=((√π)/2),Γ((5/2))=((3(√π))/4),Γ((7/2))=((15(√π))/8) ⇒Γ(2+(7/2))=Γ(5+(1/2))=((3.5.7.9)/2^5 )(√π)=((945(√π))/(32)) ⇒B(2,(7/2))=((1.((15(√π))/8))/((945(√π))/(32)))=((12)/(189))=(4/(63)) B(2,(5/2))=((Γ(2).Γ((5/2)))/(Γ(2+(5/2))))=((1.((3(√π))/4))/(Γ(4+(1/2))))=(((3(√π))/4)/((3.5.7(√π))/(16)))=(4/(35)) B(2,(3/2))=((Γ(2).Γ((3/2)))/(Γ(2+(3/2))))=(((√π)/2)/(Γ(3+(1/2))))=(((√π)/2)/((3.5(√π))/8))=(4/(15)) B(2,(1/2))=((Γ(2).Γ((1/2)))/(Γ(2+(1/2))))=((√π)/((3(√π))/4))=(4/3). Therefore,F=(4/(63))+60.(4/(35))+240.(4/(15))+64.(4/3) =((9844)/(63))](Q100127.png)
Commented bymathmax by abdo last updated on 25/Jun/20
Question Number 100087 by mathmax by abdo last updated on 24/Jun/20 | ||
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Commented bybemath last updated on 25/Jun/20 | ||
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Answered by 1549442205 last updated on 26/Jun/20 | ||
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Commented bymathmax by abdo last updated on 25/Jun/20 | ||
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