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Wondering if you can help me with this one. Part (a) asks by completing the square (and then doing a trigonometric substitution),
1/2
∫dx/x^2-x+1=π/3√3
0
(b) by factoring x^3 +1 as a sum of cubes, rewrite the integral in part (a). Then express 1/x^3+1 as the sum of a power series and use it to prove the following formula for π
∞
Π=3√3/4Σ(-1)”/8”[2/3n=1 + 1/3n=2]
N=0
½
∫dx/(x²-x+1) = ?
0
We know that (x²-x+1) = ¼[ (2x-1)²/3 + 1], which also means :
(x²-x+1) = ¼*3*( [(2x-1)/√3]² + 1 ). Thus,
1/(x²-x+1) = 4*⅓ /( [(2x-1)/√3]² + 1 ) .
Which also can be written as :
1/(x²-x+1) = (2/√3)*(2/√3) / ( [(2x-1)/√3]² + 1 ) .
Note that (2/√3) / ( [(2x-1)/√3]² + 1 )
is exactly d/dx of Arctan[(2x-1)/√3]. Hence,
∫dx/(x²-x+1) = (2/√3)*Arctan[(2x-1)/√3].
Now, let's substitute the upper & lower limits:
½
∫dx/(x²-x+1) = (2/√3)*{ Arctan[(2*½-1)/√3] - Arctan[(2*0-1)/√3] }.
0
=(2/√3)* { Arctan[0]-Arctan[-1/√3] } = π/(3√3) .
---------------------------------------------
As for part b , it's not clear what it asks for ..
Please refrase it again .
Thank you.
Alon.
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Answers by Expert:
Alon Mandes
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Kind of questions I can answer : Limits, Derivatives, Integration, Implicit functions, continuousity, differentiation ,Extremum problems, Lagrange multipliers, Gradients, Surface integrals, Multi variables functions ,Multi variables Integrals,Complex variables ,Complex functions, Curves, Trajectory integrals & Vector analyse,Divergence,Rotor & word problems. Kind of question I can't answer : Economics,Combinatorics,infinite series & convergence ,Statistics & Probabilities .
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Mathematics & Physics.
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M.A in Mathematics & Bs.c in Electronics.
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