How do you find the integral of #int cos^n(x)# if m or n is an integer?

Question

Emma Aldridge · Accepted Answer

See the explanation for one way to do these.

Even #n# For #n=2k# use #cos(2x) = 2cos^2x-1# to get the power reduction identity: #cos^2x = 1/2(1+cos(2x))# So #(cos^2x)^k = (1/2(1+cos(2x)))^k# Expand the power #k# and reduce any Odd #n# For #n=2k+1# rewrite as #cos^(2k+1)x = (cos^2x)^k cosx# # = (1-sin^2x)^k cosx# Now use # cos(2x) = 1-2sin^2x# to get the power reduction identity: #sin^2x = 1/2(1-sin(2x))#. Us power reduction, expanding the power #k#, then power reduction again and repeat as needed to get an integral of the form #int(k+a_nsin^(b_n) u + a_(n-1)sin^(b_2) u + * * * +a_1sin^(b_1)u) cosu du# Next, use substitution to integrate each term one by one.

Elijah Abram · Answer

undefined To find the integral of $\int \cos^n(x) \, dx$ where $n$ is a positive integer, you can use the method of reduction formulas. The formula for $n > 1$ is:

$$ \int \cos^n(x) \, dx = \frac{1}{n} \cos^{n-1}(x) \sin(x) + \frac{n-1}{n} \int \cos^{n-2}(x) \, dx $$

And for $n = 1$:

$$ \int \cos(x) \, dx = \sin(x) + C $$

Where $C$ is the constant of integration. You can repeatedly apply the reduction formula until the integral becomes manageable.