Solution
For the integral ∫xcos(2x)dx, use integration by parts ∫udv=uv−∫vdu.
Let u=x and dv=cos(2x)dx.
Then du=(x)′dx=1dx (steps can be seen ») and v=∫cos(2x)dx=2sin(2x) (steps can be seen »).
The integral becomes
∫xcos(2x)dx=(x⋅2sin(2x)−∫2sin(2x)⋅1dx)=(2xsin(2x)−∫2sin(2x)dx)
Apply the constant multiple rule ∫cf(x)dx=c∫f(x)dx with c=21 and f(x)=sin(2x):
2xsin(2x)−∫2sin(2x)dx=2xsin(2x)−(2∫sin(2x)dx)
Let u=2x.
Then du=(2x)′dx=2dx (steps can be seen »), and we have that dx=2du.
Therefore,
2xsin(2x)−2∫sin(2x)dx=2xsin(2x)−2∫2sin(u)du
Apply the constant multiple rule ∫cf(u)du=c∫f(u)du with c=21 and f(u)=sin(u):
2xsin(2x)−2∫2sin(u)du=2xsin(2x)−2(2∫sin(u)du)
The integral of the sine is ∫sin(u)du=−cos(u):
2xsin(2x)−4∫sin(u)du=2xsin(2x)−4(−cos(u))
Recall that u=2x:
2xsin(2x)+4cos(u)=2xsin(2x)+4cos((2x))
Therefore,
∫xcos(2x)dx=2xsin(2x)+4cos(2x)
Add the constant of integration:
∫xcos(2x)dx=2xsin(2x)+4cos(2x)+C