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How to you solve this indefinite integral from 0 to 1? ∫(xsin(πx^2))dx?
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How to you solve this indefinite integral from 0 to 1? ∫(xsin(πx^2))dx?
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Answer #1
| 31/05 2017 06:26
u = pi * x^2 du = 2 * pi * x * dx (1/(2pi)) * du = x * dx sin(pi * x^2) * x * dx => sin(u) * (1/(2pi)) * du => (1/(2pi)) * sin(u) * du Integrate (-1/(2pi)) * cos(u) + C => (-1/(2pi)) * cos(pi * x^2) + C From 0 to 1 (-1/(2pi)) * cos(pi * 1^2) - (-1/(2pi)) * cos(pi * 0^2) => (1/(2pi)) * (cos(0) - cos(pi)) => (1/(2pi)) * (1 - (-1)) => (1/(2pi)) * (1 + 1) => 2 / (2pi) => 1/pi
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Answer #1| 31/05 2017 06:2650 %