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

#### Risposta:

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

#### Spiegazione:

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!