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

##### 1 Answer

#### Answer

#### Answer:

#### Explanation

#### Explanation:

#### 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

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

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

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

**One more method using simplification:**

We will use the identity

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

From here, let

#=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

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