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.2x^5+x+C
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How do you integrate cot3x dx?
1/3ln(sin3x) + C
How do you integrate f dx?
f dx ?? do you mean f df ? int(f df) (1/2)f2 + C --------------
How do you integrate e power minus 2x?
∫e^(-2x) dx Let u = 2x du= 2 dx dx=(1/2) du ∫e^(-2x) dx = (1/2) ∫e^-u du = (1/2) (-e^-u) = -e^-u /2 + C = -e^-(2x) / 2 + C
How do you integrate cotxdx?
The integral of cot(x)dx is ln|sin(x)| + C
Double integral of x siny dx dy?
We have:int int (x * sin(y)) dx dyIntegrate x first:int(x)dx = 1/2 * x2 + CNow integrate sin(y):int(sin(y))dy = -cos(y) + CMultiply:-1/2 * x2 * cos(y) + C
How do you integrate cot3x dx?
1/3ln(sin3x) + C
How do you Integrate the following S4x3-2x2 x-1dx in Calculus?
∫(4x3 - 2x2 + x - 1) dx You can integrate this by taking the antiderivative of each term. Each of these terms is in the format axn, the antiderivative of which is axn-1/n: = ∫(4x3)dx - ∫(2x2)dx + ∫(x)dx - ∫(1)dx = x4 - 2x3/3 + x2/2 - x + C
How do you integrate f dx?
f dx ?? do you mean f df ? int(f df) (1/2)f2 + C --------------
How do you integrate sin7x?
so the problem is ∫(sin7x)dx you have to use u-substitution, set u=7x then du/dx=7 so du=7dx and solve for dx 1/7du=dx, so you get 1/7∫(sinu)du then integrate like normal, and you get 1/7(-cosu)+C then you plug u back in and get 1/7(-cos7x)+C
How do you integrate square root of 1 over x-1 dx?
2
Integrate x 5x dx?
integrate(x5x dx) simplifies to integrate(5x^2 dx), and using the power rule of integration, add one to the power of x and divide the term by that number. Thus, x5x dx integrated is (5/3)x^3
How do you integrate x power -1?
x-1 = 1/x ∫1/x dx = ln x + C
How do you integrate 1 over x-1?
2
Integrate square root of 4 xdivide by x?
∫[√(4x) / x] dx = ∫(2 / √x)dx = 2∫(x-1/2) dx = 2(2x1/2 + C) = 4√x + C
How do you integrate cosine squared times sine?
Trying to integrate: cos2x sin x dx Substitute y = cos x Then dy = -sin x dx So the integral becomes: -y2dy Integrating gives -1/3 y3 Substituting back: -1/3 cos3x
How do you integrate e power minus 2x?
∫e^(-2x) dx Let u = 2x du= 2 dx dx=(1/2) du ∫e^(-2x) dx = (1/2) ∫e^-u du = (1/2) (-e^-u) = -e^-u /2 + C = -e^-(2x) / 2 + C
Integral of 6x to the third power plus 8 all to the 4th times 18x squared dx?
This can easily be integrated by substitution. Integrate[(6x3+8)418x2 dx] let u=(6x3+8) du/dx= 18x2 du=18x2 dx Substituting u in gives... Integrate [u4 du] = (1/5)(u5) + C Substituting x back in gives.. (1/5)(6x3+8)5 + C
