Question

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

Answer 1

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)`.