How do I evaluate the indefinite integral #intcos^5(x)dx# ?
Solving problems like these is often a matter of breaking up the integral using trig identities. The Pythagorean identity (
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#
#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
We will let
Rewriting the above integrals in terms of
#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
#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
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