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I will assume that we are taking d/dx, not d/dn. There are two ways to interpret what you asked. First way is (sinx)^(2n). Second way is sin(x^(2n)).
First answer: 2n(sinx)^(2n-1)(cosx)=2ncosx(sinx)^(2n-1).
Second answer: cos(x^(2n))(2nx^(2n-1)).
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∙ 10y ago
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How do you solve sin2x plus sinX equals 0?
sin(2x) + sin(x) = 0 2sin(x)cos(x) + sin(x) = 0 sin(x)[2cos(x) + 1] = 0 sin(x) = 0 OR 2cos(x) + 1 = 0 sin(x) = 0 OR cos(x) = -1/2 x = n*pi OR x = 2/3*pi + 2n*pi OR x = -2/3*pi + 2n*pi x = pi*[2n + (0 OR 2/3 OR 1 OR 4/3)] Note that n may be any integer. The solutions in [-2pi, 2pi] are: -2pi, -4/3pi, -pi, -2/3pi, 0, 2/3pi, pi, 4/3pi, 2pi
How do you integrate sinx divide sinx plus cosx?
1/2(x-ln(sin(x)+cos(x)))
How do you find the general solutions of csc x equals -2?
Cosec x = -2 => sin x = -0.5 The primary solution is x = -pi/6 radians. Therefore the solutions are: 2n*pi - pi/6 and (2n+1)*pi + pi/6 for all integer n.
Using the remainder formula for the Taylor polynomial approximation estimate the error in your approximation to sin n?
If you use n terms from the Taylor expansion, the absolute value of the error is less than [|x|^(2n+1)]/(2n+1)!If you use n terms from the Taylor expansion, the absolute value of the error is less than [|x|^(2n+1)]/(2n+1)!If you use n terms from the Taylor expansion, the absolute value of the error is less than [|x|^(2n+1)]/(2n+1)!If you use n terms from the Taylor expansion, the absolute value of the error is less than [|x|^(2n+1)]/(2n+1)!
What would be the integral of x where x has power 2?
If x has the power 2 then you want the integral of x2, I think. When you integrate this you get : x3/3 , plus a constant.
How do you integrate cotxdx?
The integral of cot(x)dx is ln|sin(x)| + C
How find sin theta?
Sin (theta) can most easily be found on a scientific calculator. You can also approximate it with Taylor's Series... sin(x) = SummationN=0toInfinity (-1N / (2N + 1) !) (x(2N+1))) sin(x) = x - x3/3! + x5/5! - x7/7! + ... Using only the four terms above, you can approximate sin(x) within about 0.000003 in the interval x = [-1, +1].
How do you solve sin2x plus sinX equals 0?
sin(2x) + sin(x) = 0 2sin(x)cos(x) + sin(x) = 0 sin(x)[2cos(x) + 1] = 0 sin(x) = 0 OR 2cos(x) + 1 = 0 sin(x) = 0 OR cos(x) = -1/2 x = n*pi OR x = 2/3*pi + 2n*pi OR x = -2/3*pi + 2n*pi x = pi*[2n + (0 OR 2/3 OR 1 OR 4/3)] Note that n may be any integer. The solutions in [-2pi, 2pi] are: -2pi, -4/3pi, -pi, -2/3pi, 0, 2/3pi, pi, 4/3pi, 2pi
How do you integrate sinx divide sinx plus cosx?
1/2(x-ln(sin(x)+cos(x)))
How do you integrate x power x?
e^x/1-e^x
How do you integrate 2sinxcosx?
Integrate 2sin(x)cos(x)dxLet u = cos(x) and du = -sin(x)dx and pull out the -2:-2[Integral(u*du)]Integrate with respect to u:-2(u2)/2 + CSimplify:-u2 + CReplace u with cos(x):-cos2(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
Integrate sin x cos 2x?
By Angle-Addition, cos(2x) = 2cos(x)^2-1 So, sin(x)cos(2x) = [2cos(x)^2-1]sin(x) = 2sin(x)cos(x)^2 - sin(x) Int[2sin(x)cos(x)^2 - sin(x)] = (-2/3)cos(x)^3 + cos(x) +K
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
What is the antiderivative of -cscxcotx?
First, antiderivative = a solution to the indefinite integral therefore to integrate -(csc(x))(cot(x)) first convert it to -cos(x)/sin2(x) To integrate ∫-cos(x)/sin2(x) dx, use substitution u = sin(x) and du/dx = cosx This will make it ∫-1/u2 du and the antiderivative is 1/u +c, therefore the answer is 1/sin(x) + c.
How do you integrate x power -1?
x-1 = 1/x ∫1/x dx = ln x + C
What is answer of x to the power into sine x?
I've never seen the word "into" in this context. If it means x to the power of sin(x) then it's xsin(x) and for any x you'll have to look up sin (x) and that is the power you have to raise x by.
