Question

How do I evaluate the indefinite integral `intcos^5(x)dx` ?

Answer 1

Solving problems like these is often a matter of breaking up the integral using trig identities. The Pythagorean identity (`sin^2 x + cos^2 x = 1`) will be heavily involved.

Let us begin.

`int cos^5 x dx =` ?

First, note that this statement is equivalent to the statement:

`int cos^3 x * cos^2 x dx`

And, due to the Pythagorean identity, we know that

`cos^2 x = 1 - sin^2 x`

So, we will substitute:

`int cos^3 x * (1 - sin^2 x) dx`

Again, note that this statement is equivalent to the statement:

`int cos x * cos^2 x * (1 - sin^2 x) dx`

We will substitute once more:

`int cos x * (1 - sin^2 x) * (1 - sin^2 x) dx`

Now, we will simply FOIL the two sine binomials:

`int cos x * (1 - 2sin^2 x + sin^4 x) dx`

Distributing the `cos x` yields:

`int (cosx - 2sin^2 x cosx + sin^4 x cos x) dx`

Now, we will break this integral up into multiple integrals.

`int cosx dx - int 2sin^2 x cosx dx + int sin^4 x cos x dx`

The first integral is rather easy to get out of the way:

`sin x - int 2sin^2 x cosx dx + int sin^4 x cos x dx`

At this point, we have sufficiently broken up the integral. The solution can now be found by `u`-substitution.

We will let `u = sin x`. Therefore, `du = cos x dx`

Rewriting the above integrals in terms of `u` gives us:

`sin x - 2 int u^2 du + int u^4 du`

Power rule takes care of the rest. Remember the constant of integration:

`sin x - (2u^3)/ 3 + u^5 / 5 + C`

And now, we simply substitute back for `u`, and get our solution.

`int cos^5 x dx = sin x - (2sin^3 x)/ 3 + (sin^5 x)/ 5 + C`

So, in summary, when you have integrals involving odd powers of `sin x` or `cos x`, the Pythagorean identity can be used to break the integrals up to the point where they can be easily solved by `u`-substitution.