Integrate Sin(x)/[cos^2(x)-3cos(x)-10]

dyBAR25

From: US
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Votes: 30
Integrate Sin(x)/[cos^2(x)-3cos(x)-10]

 

Answer No.1

Orbaku Jackson

From: US
Posts: 10
Votes: 10
U = cos(x), -sin(x) dx = du, so sin(x) dx = -du cos^2(x) - 3cos(x) - 10 = u^2 - 3u - 10 = (u - 5)(u + 2) integral -du/[(u - 5)(u + 2)] = integral -du*[1/(u - 5) - 1/(u + 2)]*(1/7) = (-1/7) [ln|u - 5|- ln|u + 2|] + C = (-1/7) [ln|cos(x) - 5| - ln|cos(x) + 2|] + C

Answer No.2

Anonymous

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I = sin(x)/[cos^2(x)-3cos(x)-10] dx Let u = cos(x) -> du = - sin(x)dx I = (-1)/(u^2-3u-10)du = (-1)/[(u-5)(u+2)]du = A/(u-5)du + B/(u+2)du A/(u-5) + B/(u+2) = [A(u+2)+B(u-5)]/[(u-5)(u+2)] = [(A+B)u + 2A - 5B]/[(u-5)(u+2)] =>(A+B)u + 2A - 5B = - 1 =>A+B = 0 and 2A - 5B = - 1 B = -A --> 7A =-1 --> A = -1/7 and B = 1/7 Therefore I = (-1/7)/(u-5)du + (1/7)/(u+2)du I = -(1/7)Ln|u-5| + (1/7)Ln|u+2| + C I = (1/7)(Ln|u+2| - Ln|u-5|) + C I = (1/7)Ln|(u+2)/(u-5)| + C I = (1/7)Ln|(cos(x)+2)/(cos(x)-5)| + C

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