∫ cot(x) dx is written as:
∫ cos(x) / sin(x) dx
Let u = sin(x). Then, du = cos(x) dx, giving us:
∫ 1/u du
So the integral of 1/u is ln|u|. So the answer is ln|sin(x)| + c
- ln (cscx + cotx) + C You use u substitution.
The integral of cot (x) dx is ln (absolute value (sin (x))) + C. Without using the absolute value, you can use the square root of the square, i.e. ln (square root (sin2x)) + C
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Do you mean the Convolution Integral?
The integral of -x2 is -1/3 x3 .
- ln (cscx + cotx) + C You use u substitution.
The integral of cot (x) dx is ln (absolute value (sin (x))) + C. Without using the absolute value, you can use the square root of the square, i.e. ln (square root (sin2x)) + C
The derivative of cot(x) is -csc2(x).
There is no sensible or useful simplification.
Do you mean? d/dx(cotX) = - csc2X --------------
(tanx+cotx)/tanx=(tanx/tanx) + (cotx/tanx) = 1 + (cosx/sinx)/(sinx/cosx)=1 + cos2x/sin2x = 1+cot2x= csc2x This is a pythagorean identity.
(1 - csc2x)/(sinx*cotx) = -cot2x/sinxcotx = -cotx/sinx = -(cosx/sinx)/sinx = -cosx/sin2x = -cosx/(1-cos2x) = cosx/(cos2x - 1)
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Integral in Tagalog: mahalaga
In reimann stieltjes integral if we assume a(x) = x then it becomes reimann integral so we can say R-S integral is generalized form of reimann integral.
y = (sinx)^(e^x) ln(y) = ln((sinx)^(e^x)) ln(y) = (e^x)ln(sinx) (1/y)dy = (e^x)(1/sinx)(cosx)+ln(sinx)(e^x)dx (1/y)dy = (e^x)(cotx)+ln(sinx)(e^x)dx dy = ((sinx)^(e^x))((cotx)(e^x)+ln(sinx)(e^x))dx dy = ((e^x)(sinx)^(e^x))(cotx+ln(sinx))dx
Elections are integral to democracies.