Question

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

Answer 1

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

Explanation:

Use the identity `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`

The antiderivative of `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`

Let `u = cosx`. Then `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`, where `C` is a constant.

`=-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`

Hopefully this helps!