To solve the integral ∫xsin(2x)dx using Integration by Parts (IBP), we follow these steps:
Step 1: Identify parts for Integration by Parts (IBP)
Recall that Integration by Parts is given by:
∫udv=uv−∫vdu
For the integral ∫xsin(2x)dx, we choose:
- u=x (since its derivative simplifies)
- dv=sin(2x)dx (which we will integrate)
Step 2: Compute du and v
Differentiate u and integrate dv:
du=dx
v=∫sin(2x)dx
To integrate v, let’s compute ∫sin(2x)dx:
- Use substitution: Let w=2x, hence dw=21dx.
∫sin(2x)dx=2∫sin(w)dw=−2cos(w)+C=−2cos(2x)
So, v=−2cos(2x).
Step 3: Apply the Integration by Parts formula
Now, apply the formula:
∫xsin(2x)dx=u⋅v−∫vdu
Substitute the values of u, v, and du:
∫xsin(2x)dx=x⋅(−2cos(2x))−∫(−2cos(2x))dx
Simplify:
=−2xcos(2x)+2∫cos(2x)dx
Step 4: Integrate the remaining integral
Integrate ∫cos(2x)dx:
- Using the same substitution w=2x, dw=21dx:
∫cos(2x)dx=2∫cos(w)dw=2sin(w)+C=2sin(2x)
So the remaining integral is:
2∫cos(2x)dx=4sin(2x)
Step 5: Write the final answer
Combining all parts:
∫xsin(2x)dx=−2xcos(2x)+4sin(2x)+C
where C is the constant of integration.
Would you like any details on these steps, or do you have any questions?
Here are 5 related questions for further practice:
- Integrate ∫xcos(2x)dx using Integration by Parts.
- Evaluate ∫xe2xdx using Integration by Parts.
- Solve the integral ∫x2sin(x)dx using Integration by Parts.
- Integrate ∫sin(3x)dx using substitution.
- Find the integral ∫x2exdx using Integration by Parts.
Tip: When applying Integration by Parts, always aim to simplify the integrand with your choice of u and dv.