Integral of [1/(sin x cos x) dx] (substitute sin2 x + cos2 x for 1)= Integral of [(sin2 x + cos2 x)/(sin x cos x) dx]= Integral of [sin2 x/(sin x cos x) dx] + Integral of [cos2 x/(sin x cos x) dx]= Integral of (sin x/cos x dx) + Integral of (cos x/sin x dx)= Integral of tan x dx + Integral of cot x dx= ln |sec x| + ln |sin x| + C
To find the integral, you can split the integral of 3sinx-5cosx into two separate integrals. For the integral of 3sinx dx integral of sinx= -cosx you can pull the three out of the integral as a coefficient Therefore, integral of 3sinx dx= -3 cosx +C For the integral of -5cosx dx integral of cosx=sinx you can pull the -5 out of the integral as a coefficient Therefore the integral of -5cosxdx= -5sinx +C Therefore the integral of 3sinx-5cosx =-3cosx-5sinx + C
The integral of x cos(x) dx is cos(x) + x sin(x) + C
An indefinite integral is a version of an integral that, unlike a definite integral, returns an expression instead of a number. The general form of a definite integral is: ∫ba f(x) dx. The general form of an indefinite integral is: ∫ f(x) dx. An example of a definite integral is: ∫20 x2 dx. An example of an indefinite integral is: ∫ x2 dx In the definite case, the answer is 23/3 - 03/3 = 8/3. In the indefinite case, the answer is x3/3 + C, where C is an arbitrary constant.
integral (a^x) dx = (a^x) / ln(a)
Integral[sin(x)cos(x)sin2(x)cos3(x)] dxgather termsintegral[sin3(x) cos4(x)] dxpull one sin(x) as sin is oddintegral[sin2(x) cos4(x) sin(x)] dxusing trig identitiesintegral[(1 - cos2(X)) cos4(x) sin(X)] dxu substitutionu = cos(x)du = - sin(x) dxsointegral[(1 - u2)) (u4) - du] dx- integral[(1- u2)) u4 du] dx= u - 1/3u3 + 1/5u5 du= cos(x) - 1/3cos3(x) + 1/5cos5(x) + C============================
S sin3X dx =-1/3 cos3X
∫ cos(x) dx = -sin(x) + C
The integral of arcsin(x) dx is x arcsin(x) + (1-x2)1/2 + C.
Integral (14+x^4) dx = 14x + x^5/5 + C
By antiderivative do you mean integral? If yes, integral x^1 dx= (x^2)/2
Integral of 2x dx /(2x-2) Let 2x=u 2 dx = du dx = (1/2) du Integral of 2x dx /(2x-2) = Integral of (1/2) u du / (u-2) = Integral of 1/2 [ (u-2+2) / (u-2)] dx = Integral of 1/2 [ 1+ 2/(u-2)] dx = u/2 + (1/2) 2 ln(u-2) + C = u/2 + ln(u-2) + C = (2x/2) + ln(2x-2) + C = x + ln(2x-2) + C
integral x/(x-1) .dx = x - ln(x-1) + c where ln = natural logarithm and c = constant of integration alternatively if you meant: integral x/x - 1 .dx = c
âˆ« af(x) dx = a âˆ« f(x) dx
The indefinite integral of (1/x^2)*dx is -1/x+C.
∫ -3x dx = -3/2 x2 + c
Yes, the integral of gx dx is g integral x dx. In this case, g is unrelated to x, so it can be treated as constant and pulled outside of the integration.
d/dx âˆ« f(x) dx = f(x)
If you mean integral SQRT(4-x2)dx, 2arcsin(.5x)+.5xSQRT(4-x2).
If df(x)/dx = g(x), then integral [from a to b] g(x) dx = f(b)-f(a). In plain English: the definite integral can be calculated by finding the antiderivative, evaluating it at the endpoints, and subtracting.
âˆ« f(x) +g(x) dx = âˆ« f(x) dx + âˆ« g(x) dx.
The integral of cot(x)dx is ln|sin(x)| + C
See related link below for answer
Integral( sin(2x)dx) = -(cos(2x)/2) + C
The definite integral of the function f=x*exp(k*x) is (1/k)*(x-(1/k))*exp(k*x). So you have the answer to your questions by setting k equal to 1 then 2. I derived my formula by using integration by parts, setting u=x and dv=exp(k*x)dx.