How do you integrate #intx cos 2x dx#?

Question

Isabella Ackerman · Accepted Answer

#intxcos(2x)dx=# by parts:
#=xsin(2x)/2-[intsin(2x)/2dx]=#
#=xsin(2x)/2+1/4cos(2x)+c#

Charles Anctil · Answer

Make use of integration by parts

#\intu \quad dv=uv-intv du#

Let #u=x#, #\quad \implies du=dx#

and let #\quad \quaddv=cos(2x)dx#, #\implies v=1/2sin(2x)#

Integrate now by sections.

#intxcos(2x)dx=intu\quaddv=uv-intvdu#

#\quad \quad \quad \quad \quad \quad \quad \quad\quad \quad \quad \quad\quad \quad \quad \quad\quad \quad \quad \quad =x*1/2sin(2x)-int1/2sin(2x)dx#

#\quad \quad \quad \quad \quad \quad \quad \quad\quad \quad \quad \quad\quad \quad \quad \quad\quad \quad \quad \quad =x/2sin(2x)+1/4cos(2x)+C#

where #C# is the constant of integration.

A brief note on the formula for integration by parts:

The differential of #uv# is

#d[uv]=udv+vdu#

#udv=d[uv]-vdu#

Combine the two sides

#\int udv=int d[uv]-int vdu#

#\int u dv=uv-intvdu#

Aurora Alvarado · Answer

undefined To integrate $ \int x \cos(2x) \, dx $, you can use integration by parts:

Let $ u = x $ and $ dv = \cos(2x) \, dx $.
Then, $ du = dx $ and $ v = \frac{1}{2} \sin(2x) $.

Now, use the integration by parts formula:
$$ \int u \, dv = uv - \int v \, du $$

Plugging in the values:
$$ \int x \cos(2x) \, dx = \frac{1}{2}x \sin(2x) - \frac{1}{2} \int \sin(2x) \, dx $$

Integrating $ \int \sin(2x) \, dx $ gives:
$$ -\frac{1}{4} \cos(2x) + C $$

Therefore, the final result of $ \int x \cos(2x) \, dx $ is:
$$ \frac{1}{2}x \sin(2x) - \frac{1}{4} \cos(2x) + C $$
where $ C $ is the constant of integration.