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Question
0 sin w
S ------------
-pi/2 (3+2cosw)^2
evaluate the integral where (s) is the sign of integration and 0 and pi/2 are the limits.
I need my answer in detail.Thank You
In this integral we will use the deriving rule : [1/f]'=-f'/f².
In our case f=3+2cos(w). Therefore f'=-2sin(w). That means :
[ 1/(3+2cosw) ]'= 2sinw / (3+2cosw)². Now let's calculate our
integral :
∫sinw/(3+2cosw)² dw =(½)∫2sinw/(3+2cosw)² dw = (½)*1/(3+2cosw) =
1/(6+4cosw). Where w goes from -π/2 to 0 , So we get :
Integral = 1/(6+4cos[0]) - 1/(6+4cos[-π/2]) = (1/10) - (1/6) = -1/15
Alon.
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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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