d/dx cosec(x) = - cosec(x) * cot(x) so the second derivative or d(d/dx)/dx cosec(x) = [- cosec(x) * d/dx cot(x)] + [ - d/dx cosec(x) * cot(x)] = [- cosec(x) * -cosec^2(x)] + [ - (- cosec(x) * cot(x)) * cot(x)] = cosec(x) * cosec^2(x) + cosec(x)*cot^2(x) = cosec(x) * [cosec^2(x) + cot^2(x)].
.5(x-sin(x)cos(x))+c
the integral of the square-root of (x-1)2 = x2/2 - x + C
(x+sinxcosx)/2, can do it by parts or by knowing your double angle formulas
Cot x is 1/tan x or cos x / sin x or +- sqrt cosec^2 x -1
∫ cos(x)/sin2(x) dx = -cosec(x) + C C is the constant of integration.
d/dx cosec(x) = - cosec(x) * cot(x) so the second derivative or d(d/dx)/dx cosec(x) = [- cosec(x) * d/dx cot(x)] + [ - d/dx cosec(x) * cot(x)] = [- cosec(x) * -cosec^2(x)] + [ - (- cosec(x) * cot(x)) * cot(x)] = cosec(x) * cosec^2(x) + cosec(x)*cot^2(x) = cosec(x) * [cosec^2(x) + cot^2(x)].
integral of radical sinx
The indefinite integral of (1/x^2)*dx is -1/x+C.
arctan(x)
maths signs
.5(x-sin(x)cos(x))+c
the integral of the square-root of (x-1)2 = x2/2 - x + C
Yes of course cosec x is the inverse of sin x by definition in trigonometry sin x=opp. side/hypotenuse cosec x= hypotenuse/opp.side thank u
tan(sqrtX) + C
(x+sinxcosx)/2, can do it by parts or by knowing your double angle formulas
ln(sinx) + 1/3ln(sin3x) + C