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To integrate such a function, you must use the u-substitution technique. Let u = 4x; therefore du = 4 dx. Then du/4 = dx. The integral becomes : Int(cos 4x dx) = Int[(cos u)(du/4)] = ¼ Int(cos u du) = ¼ (sin u) + C = ¼ sin 4x + C
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What is the derivative of ddx(a2x)?
Assuming you mean what is the value of the derivative d/dx(a²x), then: d/dx(a²x) = a² The derivative (with respect to x) of d/dx(a²x) = d/dx(d/dx(a²x)) = d/dx(a²) = 0.
What is the derivative of ln 1 divided by 1-x?
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Integral of 1 divided by sinx cosx?
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
What is derivative of 1 - x?
d/dx (1-x)=d/dx (1)-d/dx(x)=0-1=-1
What is the derivative of x over 1 plus x squared?
The product rule states: d/dx uv = vdu/dx + udv/dxIt is not quite clear what your denominator is:"over 1 + (x squared)": d/dx (x/1+x2)d/dx(x/1+x2) = d/dx x(1 + x2)-1 = (1 + x2)-1 d/dx x + x d/dx (1 + x2)-1= (1 + x2)-1 + x -2x (1 + x2)-2= (1 + x2)-2 (1 + x2 - 2x2)= (1 - x2) / (1 + x2)2"over (1 + x) [all] squared": d/dx (x/(1+x)2)d/dx(x/(1+x)2) = d/dx x(1 + x)-2 = (1 + x)-2 d/dx x + x d/dx (1 + x)-2= (1 + x)-2 + x -2 (1 + x)-3= (1 + x)-3 (1 + x - 2x)= (1 - x)/(1 + x)3If you prefer, you can use the quotient rule: d/dx (u/v) = (v du/dx - udv/dx) / v2
Integral of sinx cosx sin2x cos3x dx?
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============================
How do you differentiate 5 sqrtx dx?
( no dx needed as you are taking the differential, not integrating. )d/dx[5sqrt(X)]rewrite as...,d/dx[5X1/2]use power rule [ nX(n - 1)]= (5/2)X -1/2==========
What is the integral of -e to the -x?
Know that ∫eu du = eu du/dx + c ∫-e-x dx = e-x + c But ∫eu du = eu + c Perhaps we are integrating -(e-x ) though the question might be (-e)-x Question is not clear.
What is horzontal?
Horizontal integration is where the slices are parallel to the x-axis, instead of to the y-axis. In this case, you would be integrating f(y)dy, instead of f(x)dx.
What is horzontal integration?
Horizontal integration is where the slices are parallel to the x-axis, instead of to the y-axis. In this case, you would be integrating f(y)dy, instead of f(x)dx.
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
What is the general solution of y''-2y'-3y=e^x?
1
How can you integrate a three variable functions?
To integrate a three-variable function, you need to perform a triple integral. This involves integrating the function with respect to each variable individually, in the order specified by the given problem. The result will be a scalar value representing the integral of the three-variable function over the specified region.
What is the Second derivative of 3.9625.lnx?
3.9625lnx?The first derivative is:d/dx(cu)=c*du/dx where c is a constant.d/dx(3.9625lnx)=3.9625*d/dx(lnx)-The derivative of lnx is:d/dx(lnu)=(1/u)*d/dx(u)d/dx(lnx)=(1/x)*d/dx(x)d/dx(3.9625lnx)=3.9625*[(1/x)*d/dx(x)]-The derivative of x is:d/dx(xn)=nxn-1d/dx(x)=1*x1-1d/dx(x)=1*x0d/dx(x)=1*(1)d/dx(x)=1d/dx(3.9625lnx)=3.9625*[(1/x)*1]d/dx(3.9625lnx)=3.9625*(1/x)d/dx(3.9625lnx)=3.9625/xThe second derivative of 3.9625lnx is the derivative of 3.9625/x=3.9625*x-1:d/dx(cu)=c*du/dx where c is a constant.d/dx(3.9625*x-1)=3.9625*d/dx(x-1)-The derivative of x-1 is:d/dx(xn)=nxn-1d/dx(x-1)=-1*x-1-1d/dx(x-1)=-1*x-2d/dx(x-1)=-1/x2d/dx(3.9625*x-1)=3.9625*(-1/x2)d/dx(3.9625*x-1)=-3.9625/x2
What is the derivative of ddx(a2x)?
Assuming you mean what is the value of the derivative d/dx(a²x), then: d/dx(a²x) = a² The derivative (with respect to x) of d/dx(a²x) = d/dx(d/dx(a²x)) = d/dx(a²) = 0.
What is the derivative of ln 1 divided by 1-x?
0
What is the derivative of secxtanx?
d/dx(uv)=u*dv/dx+v*du/dxd/dx(secxtanx)=secx*[d/dx(tanx)]+tanx*[d/dx(secx)]-The derivative of tanx is:d/dx(tan u)=[sec(u)]2*d/dx(u)d/dx(tan x)=[sec(x)]2*d/dx(x)d/dx(tan x)=[sec(x)]2*(1)d/dx(tan x)=(sec(x))2=sec2(x)-The derivative of secx is:d/dx(sec u)=[sec(u)tan(u)]*d/dx(u)d/dx(sec x)=[sec(x)tan(x)]*d/dx(x)d/dx(sec x)=[sec(x)tan(x)]*(1)d/dx(sec x)=sec(x)tan(x)d/dx(secxtanx)=secx*[sec2(x)]+tanx*[sec(x)tan(x)]d/dx(secxtanx)=sec3(x)+sec(x)tan2(x)
