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

1 Answer
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources
preview
?

Answer

Write a one sentence answer...

Answer:

Explanation

Explain in detail...

Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

22
mason m Share
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.

Was this helpful? Let the contributor know!
1500
Trending questions
Impact of this question
17888 views around the world
You can reuse this answer
Creative Commons License