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Integration of sinx is?
-cos x + C
If y equals the integral from 5x on top to cosx on bottom of cos of u squared du what is y'?
y=S^5x _cos(x) cos(u²) du The derivative of a definite integral of a function f(x) is equal to the difference in the product of the function at each limit of integration times the limit of integration. y'=cos(u²)*du/dx from u=cos(x) to u=5x y'=-sin(x)*cos(cos(x)²)-5*cos(25x²) To understand why this works, consider the following where F(x) is the antiderivative of f(x) y=F(g(x))-F(h(x))=S f(x)dx from h(x) to g(x) If you take the derivative of this expression and apply the chain rule dy/dx = dF(g(x))/dx - dF(h(x))/dx = f(g(x))*dg/dx - f(h(x))*dh/dx
What is the answer to 5x plus 5x-10?
5x+5x-10 10x-10 is the only answer you can get from this
What is 10x minus 5x plus 5x?
10x - 5x + 5x = 10x
-10x plus 5x?
-10x + 5x = -5x
Integration of sinx is?
-cos x + C
What is the Integration of sine wave?
cos wave
Integral of (sin2x)(cos3x)?
Int[sin(2x)*cos(3x)]dx = Int[(2sinx*cosx)*(4cos^3x - 3cosx)]dx= Int[(8sinx*cos^4x - 6sinx*cos^2x)]dx Let cosx = u then du/dx = -sinx So, the integral is Int[-8*u^4 + 6*u^2]du = -8/5*u^5 + 2u^3 + c where c is a constant of integration = -8/5*cos^5x + 2cos^3x + c
If y equals the integral from 5x on top to cosx on bottom of cos of u squared du what is y'?
y=S^5x _cos(x) cos(u²) du The derivative of a definite integral of a function f(x) is equal to the difference in the product of the function at each limit of integration times the limit of integration. y'=cos(u²)*du/dx from u=cos(x) to u=5x y'=-sin(x)*cos(cos(x)²)-5*cos(25x²) To understand why this works, consider the following where F(x) is the antiderivative of f(x) y=F(g(x))-F(h(x))=S f(x)dx from h(x) to g(x) If you take the derivative of this expression and apply the chain rule dy/dx = dF(g(x))/dx - dF(h(x))/dx = f(g(x))*dg/dx - f(h(x))*dh/dx
Int cos 10x cos 15x dx?
There is some kind of formula here, half angle, or some such that I forget, but I do remember the algorithm. So...,int[cos(10X)cos(15X)] dxsince this is multiplicative, switch it aroundint[cos(15X)cos(10X)] dxint[cos(15X - 10X)/2(15 -10) + cos(15X + 10X)/2(15 + 10)] dxint[cos(5X)/10 + cos(25X)/50] dx= 1/10sin(5X) + 1/50sin(25X) + C=========================
Int sin 3x cos 5x dx?
integral sin(3 x) cos(5 x) dx = 1/16 (8 cos^2(x)-cos(8 x))+C
Integration of sin2x?
Integral( sin(2x)dx) = -(cos(2x)/2) + C
What is the integral of sin 5X cos 10X dX?
use product-to-sumformula sin u cos v =1/2 [sin(u+v)+ sin(u-v)]so you get1/2 int: (sin 15x) - (sin5x) dxsplit it1/2 int: sin 15x dx- 1/2 int: sin 5x dxusing substitution you can conclude that1/30 int: sin u du- 1/10 in sin w dw(you get the fraction change when you set dx=duand dw)so then- (cos u)/30 + (cos w)/10replace the substitution(cos 5x)/10 - (cos 15x)/30 + Constant
Why did the chicken cross the road due to the resulting effects of finding the integration of sinx in respect to cos?
Yes, he is.
Can you integrate -cos x plus c?
Why not? Just a second integration. Drop the constant. int[- cos(x)] dx the negative implies - 1 and can be brought out side the integrand - int[cos(x)] dx = - sin(x) + C ==========
Integral of xlnxdx?
integration by parts. Let u=lnx, dv=xdx-->du=(1/x)dx, v=.5x^2. Integral of (xlnxdx)=lnx*.5x^2-integral of .5x^2(1/x)dx=lnx*.5x^2-integral of .5xdx=lnx*.5x^2-(1/6)x^3. That's it.
What is the integral of the sine of x with respect to x?
∫ sin(x) dx = -cos(x) + CC is the constant of integration.
