How do you find the integral of #(sinx)(cosx)dx#?

1 Answer
Jim H
Jul 5, 2015

There are several ways to write the correct answer.

Explanation:

#intsinxcosxdx#

Solution 1
With #u = sinx# we get #sin^2x/2 +C#

Solution 2
With #u = cosx#, we get #-cos^2x/2 + c#

Solution 3
Noting that #sinxcosx = 1/2sin(2x)#, we rewrite:

#intsinxcosxdx = 1/2 int sin(2x) dx#

Now let #u = 2x# to get #1/4 cos(2x) + CC#

(I love this problem and use it every time I teach Calulus I.)

Note 1: It is not an accident that i used different notations for the constants in each solution.

Note 2:
The difference (literally -- that is, the subtraction) between the apparently different solutions are constants.
For example: #sin^2x/2 # minus #-cos^2x/2# simplifies to:

#sin^2x/2 - (-cos^2x/2) = (sin^2x+cos^2)/2 = 1/2#

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