Evaluate The Integral (x^2+1)sin2x Dx

KnowledgeMath0

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Evaluate the integral (x^2+1)sin2x dx

 

Answer No.1

LovelyProdigy

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Integral (x^2+1) sin(2 x) dx = -1/4 (2 x^2-1) cos(2 x)+1/2 x sin(2 x)-1/2 cos(2 x)+constant

Answer No.2

Anonymous

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-1/4 (2 x^2-1) cos(2 x)+1/2 x sin(2 x)-1/2 cos(2 x)+constant

Answer No.3

Anonymous

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Integrate by parts: ?udv = uv - ?vdu; ?xsin(2x)dx u = x du = 2xdx dv = sin(2x) v = -cos(2x)/2 ?xsin(2x)dx = -xcos(2x)/2 + ?x.cos(2x)dx (Integrate x.cos(2x) by parts as well) = -xcos(2x)/2 + x.sin(2x)/2 - ?sin(2x)/2 dx = -xcos(2x)/2 + x.sin(2x)/2 + cos(2x)/4 + c = cos(2x)(1/4 - x/2) + x.sin(2x)/2 + c

Answer No.4

AlphaAsian3

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Integrate by parts: ?udv = uv - ?vdu; ?xsin(2x)dx u = x du = 2xdx dv = sin(2x) v = -cos(2x)/2 ?xsin(2x)dx = -xcos(2x)/2 + ?x.cos(2x)dx (Integrate x.cos(2x) by parts as well) = -xcos(2x)/2 + x.sin(2x)/2 - ?sin(2x)/2 dx = -xcos(2x)/2 + x.sin(2x)/2 + cos(2x)/4 + c = cos(2x)(1/4 - x/2) + x.sin(2x)/2 + c

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