How do you find the integral of # sin(x)cos(x)#?

2 Answers
May 14, 2016

#int sin(x)cos(x) dx = 1/2 sin^2(x) + C#

Explanation:

Notice that using the power and chain rules we find:

#d/(dx) sin^2(x) = 2 sin(x) cos(x)#

So:

#int sin(x)cos(x) dx = 1/2 sin^2(x) + C#

May 15, 2016

This is one of my favorite integrals for introductory courses. Here are three solutions.

Explanation:

Solution 1

#I = int sinxcosx dx#

Substitute #u=sinx#, so that #du = cosx dx#.

#I = int u du = 1/2u^2 +C#.

Undo the substitution to get

#I = 1/2 sin^2x +C#

Solution 2

#I = int sinxcosx dx#

Substitute #u=cosx#, so that #du = -sinxx dx#.

#I = int -u du = -1/2u^2 +C#.

Undo the substitution to get

#I = -1/2 cos^2x +C#

Solution 3

#I = int sinxcosx dx#.

Observe that #sin2x = 2sinxcosx#, so #sinxcosx = 1/2sin(2x)#.

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

Substitute #u = 2x#, so that #du = 2 dx# and

#I = 1/4 int sinu du = 1/4 cosu +C#.

Undo the substitution to get

#I = 1/4 cos(2x) +C#

All three answers are correct.

Can you explain?

Hint: What are the differences between the answers? (In the mathematical sense of "difference".)