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

Aug 4, 2014

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.

$\int {\cos}^{5} x \mathrm{dx} =$ ?

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

$\int {\cos}^{3} x \cdot {\cos}^{2} x \mathrm{dx}$

And, due to the Pythagorean identity, we know that

${\cos}^{2} x = 1 - {\sin}^{2} x$

So, we will substitute:

$\int {\cos}^{3} x \cdot \left(1 - {\sin}^{2} x\right) \mathrm{dx}$

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

$\int \cos x \cdot {\cos}^{2} x \cdot \left(1 - {\sin}^{2} x\right) \mathrm{dx}$

We will substitute once more:

$\int \cos x \cdot \left(1 - {\sin}^{2} x\right) \cdot \left(1 - {\sin}^{2} x\right) \mathrm{dx}$

Now, we will simply FOIL the two sine binomials:

$\int \cos x \cdot \left(1 - 2 {\sin}^{2} x + {\sin}^{4} x\right) \mathrm{dx}$

Distributing the $\cos x$ yields:

$\int \left(\cos x - 2 {\sin}^{2} x \cos x + {\sin}^{4} x \cos x\right) \mathrm{dx}$

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

$\int \cos x \mathrm{dx} - \int 2 {\sin}^{2} x \cos x \mathrm{dx} + \int {\sin}^{4} x \cos x \mathrm{dx}$

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

$\sin x - \int 2 {\sin}^{2} x \cos x \mathrm{dx} + \int {\sin}^{4} x \cos x \mathrm{dx}$

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, $\mathrm{du} = \cos x \mathrm{dx}$

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

$\sin x - 2 \int {u}^{2} \mathrm{du} + \int {u}^{4} \mathrm{du}$

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

$\sin x - \frac{2 {u}^{3}}{3} + {u}^{5} / 5 + C$

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

$\int {\cos}^{5} x \mathrm{dx} = \sin x - \frac{2 {\sin}^{3} x}{3} + \frac{{\sin}^{5} x}{5} + C$

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.