HELP: Consider the indefinite integral ∫sin^3(x) dx

Question

a) Explain why the substitution u = sin (x) will not work to help evaluate the integral.

b) By observing that sin³(x) = sin(x) • sin²(x), and using the Pythagorean Identity, rewrite the integral as a product of sin(x) and another function.

c) Now explain why the substitution u = cos(x) now helps evaluate the integral.

d) Use your work in parts a), b), and c) to evaluate the integral, thereby discovering the antiderivative of sin³(x)

e) Use a similar approach to evaluate ∫ cos³(x) dx.

Answer 1

So we are asked to integrate sin³(x). Let's try to do a u-substitution to see if this will work...

∫sin³(x) * dx = ?

Let:

u(x) = sin(x)

du = cos(x) * dx

∫sin³(x) * dx = ∫u³ * du/cos(x)

As we can see this will not help us since we don't readily have an expression for cos(x) in-terms of u(x). But we can use the Pythagorean Theorem to assist us with a suitable substitution with sin²(x) + cos²(x) = 1.

∫sin³(x) * dx = ∫sin(x) * sin²(x) * dx = ∫sin(x) * (1 - cos²(x)) * dx = ∫sin(x) * dx - ∫sin(x) * cos²(x) * dx

With this original integral broken into two pieces we can attempt to solve each piece individually...

The first integral gives -cos(x) + C as the solution (with the integration constant due to the indefinite integral)

The second integral we can apply u-substitution...

Let :

u(x) = cos(x)

du = -sin(x) * dx

- ∫sin(x) * cos²(x) * dx = ∫u² du = u³/3 + C --> cos³(x)/3 + C

Putting the two parts together yields --> ∫sin³(x) * dx = -cos(x) + cos³(x)/3 + C

where we have added the two constant terms into a single unknown constant term.

Let's apply this concept to the other integral --> ∫cos³(x) * dx

∫cos³(x) * dx = ∫cos(x) * cos²(x) * dx = ∫cos(x) * (1 - sin²(x)) * dx = ∫cos(x) * dx - ∫cos(x) * sin²(x) * dx

Let u = sin(x) in the second integral

du = cos(x) * dx

∫cos³(x) * dx = -sin(x) - sin³(x)/3 + C = -(sin(x) + sin³(x)/3) + C