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cos(x) cot(x) = cos(x) * 1/(tan(x)) = cos(x) * 1 / (sinx(x) / cos(x))
= cos2(x) / sin(x) = (1-sin2(x)) / sin(x) = 1/sin(x) - sin(x)
so the antiderivative of cos(x)cot(x) = log[abs(tan(x/2))]+cos(x)
This can also be written as log[abs((sin(x)/(cos(x)+1))]+cos(x) if we want everything in terms of x and not (x/2). The two answers are, of course, the same.
where log(x) refers to the natural log, often written ln(x).
We might write ln[|sin(x)|/|cos(x)+1|] +cos(x)
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Integral of 1 divided by sinx cosx?
Integral of [1/(sin x cos x) dx] (substitute sin2 x + cos2 x for 1)= Integral of [(sin2 x + cos2 x)/(sin x cos x) dx]= Integral of [sin2 x/(sin x cos x) dx] + Integral of [cos2 x/(sin x cos x) dx]= Integral of (sin x/cos x dx) + Integral of (cos x/sin x dx)= Integral of tan x dx + Integral of cot x dx= ln |sec x| + ln |sin x| + C
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cot x = (cos x) / (sin x) cos (x - 180) = cos x cos 180 + sin x sin 180 = - cos x sin (x - 180) = sin x cos 180 - cos x sin 180 = - sin x cot (x - 180) = (cos (x - 180)) / (sin (x - 180)) = (- cos x) / (- sin x) = (cos x) / (sin x) = cot x
What is cot x?
Cot x is 1/tan x or cos x / sin x or +- sqrt cosec^2 x -1
How do you solve csc x sin x equals cos x cot x plus?
Suppose csc(x)*sin(x) = cos(x)*cot(x) + y then, ince csc(x) = 1/sin(x), and cot(x) = cos(x)/sin(x), 1 = cos(x)*cos(x)/sin(x) + y so y = 1 - cos2(x)/sin(x) = 1 - [1 - sin2(x)]/sin(x) = [sin2(x) + sin(x) - 1]/sin(x)
Prove identity cos 2xcot2 x-1cot2 x1?
the questions is 2x=(cot^2 x-1)/(cot^2 x+1)
