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Integral of sec2x-cosx plus x2dx?
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
What is the integration of tanx?
The integral of tan(x) dx = ln | sec(x) | + cto solve... tan(x) = sin(x)/cos(x)the integral of (sin(x)/cos(x) dx) ... let u = cos(x) then du = -sin(x) dx= the integral of (1/u -du)= -ln | u | + c= -ln | cos(x) | + c= ln | (cos(x))^-1 | + c ... or ... ln | 1/cos(x) | + c= ln | sec(x) | + c
What is the integral of 1 divided by the cosine of x with respect to x?
∫ 1/cos(x) dx = ln(sec(x) + tan(x)) + C C is the constant of integration.
What is sec x cos x?
sec x = 1/cos x sec x cos x = [1/cos x] [cos x] = 1
What is integration of secx tanx?
Will try integration by parts. uv - int[v du] u = sec(x)----------------du = sec(x) tan(x) dv = tan(x)---------------v = ln[sec(x)] sec(x) ln[sex(x)] - int[lnsec(x) dx] = sec(x) ln[sec(x)] - xlnsec(x) - x + C ===========================
