First, antiderivative = a solution to the indefinite integral
therefore to integrate -(csc(x))(cot(x)) first convert it to -cos(x)/sin2(x)
To integrate ∫-cos(x)/sin2(x) dx, use substitution u = sin(x) and du/dx = cosx
This will make it ∫-1/u2 du and the antiderivative is 1/u +c, therefore the answer is 1/sin(x) + c.
Antiderivative of x/-1 = -1(x^2)/2 + C = (-1/2)(x^2) + C Wolfram says antiderivative of x^-1 is log(x) + C
d/dx (-cscx-sinx)=cscxcotx-cosx
It is -exp (-x) + C.
The antiderivative, or indefinite integral, of ex, is ex + C.
The antiderivative of x2 + x is 1/3x3 + 1/2x2 + C.
The antiderivative of 2x is x2.
∫cscxcotx*dx∫csc(u)cot(u)*du= -csc(u)+C, where C is the constant of integrationbecause d/dx(csc(u))=-[csc(u)cot(u)],so d/dx(-csc(u))=csc(u)cot(u).∫cscxcotx*dxLet:u=xdu/dx=1du=dx∫cscucotu*du= -csc(u)+CPlug in x for u.∫cscxcotx*dx= -csc(x)+C
The antiderivative of a function which is equal to 0 everywhere is a function equal to 0 everywhere.
35x2
Using u-substitution (where u = sinx), you'll find the antiderivative to be 0.5*sin2x + C.
I assume you mean -10x^4? In that case, antiderivative would be to add one to the exponent, then divide by the exponent. So -10x^5, then divide by 5. So the antiderivative is -2x^5.
Antiderivative of x/-1 = -1(x^2)/2 + C = (-1/2)(x^2) + C Wolfram says antiderivative of x^-1 is log(x) + C
By antiderivative do you mean integral? If yes, integral x^1 dx= (x^2)/2
d/dx (-cscx-sinx)=cscxcotx-cosx
The fundamental theorum of calculus states that a definite integral from a to b is equivalent to the antiderivative's expression of b minus the antiderivative expression of a.
(that weird integral or antiderivative sign) x^(-6/5) dx =-5*x^(-1/5)
-e-x + C.