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evaluate integral (sin^3(x)/cos(x)) dx
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I know i have to convert sin^3(x) into sin(x) * sin^2(x) / cos(x) let u = cosx du= -sinx dx dx= -sinx/du sin^2(x) = 1-cos^2(x) so -integral = ((1-u^2) / u) du it looks like a ln function but I'm stuck from here.
(1 - u^2) / u = 1/u - u -integral = ln(u) - u^2/2 + C
integral = u2/2 - ln(u) + C = cos^2(x) / 2 - ln(cos(x)) + C
so simple. thanks!
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