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Calculus1 23 Online
OpenStudy (anonymous):

Evaluate the integral of (sin^2x -cos^2x)/(sinx) dx

OpenStudy (blurbendy):

you can rewrite this as csc(x)(sin^2(x)-cos^2(x))dx = (integral sign)(sin(x) - cos(x)cot(x))dx = (integral sign) sin(x)dx - (integral sign) cos(x)cot(x) dx = (integral sign) sin(x)dx - (integral sign) (csc(x) - sin(x)) dx = (integral sign) sin(x)dx + (integral sign) sin(x)dx - (integral sign) csc(x) dx = (integral sign) sin(x)dx + (integral sign) sin(x) dx + ln|(cot(x) + csc(x)| = - cos + (integral sign) sin(x)dx + ln|cot(x) + csc(x) = ln|cot(x) + csc(x) - 2cos(x) + C = -2cos(x) - ln|sin(x/2)| + ln|cos(x/2)| + C

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