Evaluate #intcos^5x dx# using u-substitution or tabular integration?

1 Answer
Jarob R G.
Apr 30, 2018

#sinx - 2/3 sin^3x + 1/5 sin^5x + C#

Explanation:

We will use u-substitution and a trigonometric identity.

Recall that #cos^2 x + sin^2 x = 1#. Rearranging, it follows that #cos^2 x = 1 - sin^2 x#. Note that we can use this fact to change the integral as follows:

#int cos^5 x dx = int (cos^2x)^2 cosx dx#
#= int (1-sin^2 x)^2 cosxdx#

Make a u-substitution by letting #u = sinx#. Then #du = cosx dx#. Our integral becomes:

#int (1-u^2)^2 du = int (1 - 2u^2 + u^4) du#
#= u - 2/3 u^3 + 1/5 u^5 + C#
#= sinx - 2/3 sin^3x + 1/5sin^5x + C#

This is our final answer.

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