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

1 Answer
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 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.