How do you find the integral of # Cos(2x)Sin(x)dx#?

Usa l'identità #cos(2x) = 1 - 2sin^2x#.

#=int(1 - 2sin^2x)sinxdx#

Multiply out.

#=int(sinx - 2sin^3x)dx#

Separate using #int(a + b)dx = intadx + intbdx#

#=int(sinx)dx - int(2sin^3x)dx#

L'antiderivativo di #sinx# is #-cosx#. Use the property of integrals that #int(Cf(x))dx = Cintf(x)# where #C# is a constant. Note that #sin^3x# can be factored as #sin^2x(sinx)#, which can in turn be written as #(1- cos^2x)(sinx)# by the identity #sin^2x + cos^2x = 1#.

#=-cosx - 2int(1 - cos^2x)sinxdx#

lasciare #u = cosx#. Poi #du = -sinxdx -> dx = -(du)/sinx#.

#=-cosx - 2int(1 - u^2)sinx * -(du)/sinx#

The sines under the integral cancel each other out.

#=-cosx - 2int(1 - u^2) * -(du)#

Extract the negative #1#.

#=-cosx + 2int(1 - u^2)du#

Separate the integrals.

#=-cosx + 2int1du - 2intu^2du#

Integrate using the rule #int(x^n)dx = x^(n + 1)/(n + 1) + C#, Dove #C# è una costante.

#=-cosx + 2u - 2(1/3u^(3)) + C#

Resubstitute #u = cosx# to define the function with respect to #x#.

#=-cosx + 2cosx - 2/3cos^3x + C#

Finally, combine like terms.

#=cosx - 2/3cos^3x + C#

Speriamo che questo aiuti!