What is the integral of #sin(x) cos(x)#?

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Answer 1 · 2014-12-16T05:02:06

It's #1/2 sin^2 (x) + C#.

The substitution used to solve this integral is simple.

Note that #cos(x)# is the derivative of #sin(x)#.

Define the variable #u=sin(x)#.
We have #du=cos(x)dx#.
So, #dx=1/cos(x) du#.

The integral:

#int sin(x) cos(x) dx = int u cos(x)/cos(x) du = int u du#

Knowing that #d/(du)[1/2 u^2 + C]=u# we have:

#int u du = int d/(du)[1/2 u^2 + C] du#

#int d/(du)[1/2 u^2 + C] du = 1/2 u^2 + C = 1/2 sin^2 (x) + C#