How do you integrate #sin(x)cos(x)#?

1 Answer
Aug 2, 2016

Answer:

Depending on the route you take, valid results include:

  • #sin^2(x)/2+C#
  • #-cos^2(x)/2+C#
  • #-1/4cos(2x)+C#

Explanation:

There are a variety of methods we can take:

Substitution with sine:

Let #u=sin(x)#. This implies that #du=cos(x)dx#.

Thus:

#intunderbrace(sin(x))_uoverbrace(cos(x)dx)^(du)=intudu=u^2/2+C=color(blue)(sin^2(x)/2+C#

Substitution with cosine:

Let #u=cos(x)#, so #du=-sin(x)dx#.

Therefore:

#intsin(x)cos(x)dx=-intunderbrace(cos(x))_uoverbrace((-sin(x))dx)^(du)=-intudu=-u^2/2+C#

#=color(blue)(-cos^2(x)/2+C#

Brief interlude:

You may be wondering, well, why are both of these answers valid?

Note that:

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

However, the #1/2# is absorbed into #C# as #C# represents any constant:

#=-cos^2(x)/2+C#

One more method using simplification:

We will use the identity #sin(2x)=2sin(x)cos(x)#. Thus, #sin(x)cos(x)=sin(2x)/2#.

#intsin(x)cos(x)dx=intsin(2x)/2dx=1/2intsin(2x)dx#

From here, let #u=2x# so that #du=2dx#.

#=1/4intsinunderbrace((2x))_u overbrace((2)dx)^(du)=1/4intsin(u)du=-1/4cos(u)+C= color(blue)(-1/4cos(2x)+C#

You can also show that this is equivalent to the other two answers using the identity #cos(2x)=cos^2(x)-sin^2(x)#.