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Mathematics
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int (sqrt (x^2-9))/ x x>3
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it's copied from a question before let x=3 sin(a) dx=3cos(a)da a = arcsin(x/3) int (9 sin^2(a) 3 cos(a) da / sqrt(9-9sin^2(a))) =int (27 sin^2(a)cos(a)da/sqrt(9cos^2(a)) **note:[1-sin^2(a)=cos^2(a)] = int (27 sin^2(a)cos(a)da/3cos(a)) = int (9 sin^2(a)da) **note:[sin^2(a)=sin(a)*sin(a)= 0.5(cos(0)-cos(2a)) = int (4.5(1-cos(2a))da) = int (4.5da) - int(4.5cos(2a) da) = 4.5a - 2.25sin(2a) +c =4.5arcsin(x/3) - 4.5sin(a)cos(a) + c *note:[refer to the lets;; and sin(2a)=2sin(a)cos(a)] =4.5arcsin(x/3) - 4.5(x/3)sqrt(9-x^2)/3 +c =4.5arcsin(x/3) - 0.5xsqrt(9-x^2)/9 +c
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