The integral of x cos(x) dx is cos(x) + x sin(x) + C
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The integral of arcsin(x) dx is x arcsin(x) + (1-x2)1/2 + C.
I wasn't entirely sure what you meant, but if the problem was to find the integral of [sec(2x)-cos(x)+x^2]dx, then in order to get the answer you must follow a couple of steps:First you should separate the problem into three parts as you are allowed to with integration. So it becomes the integral of sec(2x) - the integral of cos(x) + the integral of x^2Then solve each part separatelyThe integral of sec(2x) is -(cos(2x)/2)The integral of cos(x) is sin(x)The integral of x^2 isLastly you must combine them together:-(cos(2x)/2) - sin(x) + (x^3)/3
∫ 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
The integral of f'(x) = 1 is f(x) = x + c