Integral of cos^5x*sin³xdx ?

1 Answer
Jarob R G.
May 5, 2018

#-1/6 cos^6x + 1/8 cos^8x + C#

Explanation:

We wish to solve the following: #int (cos^5x sin^3x) dx#. Since #cosx# is the derivative of #sinx#, this looks ripe for a u-substitution. However, we'd like there to be only one instance of #du#, so we need to figure out a way to transform the integral so that we only have a single power of #sin# or #cos#.

Luckily, this is easy. Recall the identity #sin^2x + cos^2x = 1#. From this, it follows that #sin^2x = 1 - cos^2x#. Using this fact, we can make the following changes:

#int (cos^5x sin^3x) dx#
#= int (cos^5x sin^2x sinx) dx#
#= int((cos^5x)(1 - cos^2x)(sinx))dx#

Now, let #u = cosx#. Then #du = -sinx#. This gives

#-int ( (u^5)(1 - u^2)) du#
#= -int (u^5 - u^7) du#
#= - (1/6 u^6 - 1/8 u^8) + C#
# = -1/6 cos^6x + 1/8 cos^8x + C#

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