Evaluate the integral: I = integral x²sin2x dx
Evaluate the integral: $$ I = \int x^2\sin2x\,dx. $$
Explanation of Method
Integration by parts with $u=x^2$, $dv=\sin2x\,dx$ reduces the power of $x$ by one, leaving a $\int x\cos2x\,dx$ integral that itself requires a second application of integration by parts.
- Choose $u=x^2$, $dv=\sin2x\,dx$ and apply integration by parts.
- Solve the leftover $\int x\cos2x\,dx$ with a second integration by parts.
- Combine the results.
Step 1 · First integration by parts
Let:
$$ u = x^2 \quad \Rightarrow \quad du = 2x\,dx, \qquad dv = \sin2x\,dx \quad \Rightarrow \quad v = -\frac12\cos2x. $$
Applying the formula:
$$ I = -\frac12x^2\cos2x + \int x\cos2x\,dx. $$
Step 2 · Second integration by parts
For $\displaystyle\int x\cos2x\,dx$, let $u=x$, $dv=\cos2x\,dx$, so $du=dx$, $v=\tfrac12\sin2x$:
$$ \int x\cos2x\,dx = \frac12x\sin2x - \frac12\int \sin2x\,dx = \frac12x\sin2x + \frac14\cos2x. $$
Step 3 · Combine
Substituting back:
$$ I = -\frac12x^2\cos2x + \frac12x\sin2x + \frac14\cos2x + C. $$
Summary Table
| Piece | Value |
|---|---|
| First IBP | $-\tfrac12x^2\cos2x+\int x\cos2x\,dx$ |
| $\int x\cos2x\,dx$ | $\tfrac12x\sin2x+\tfrac14\cos2x$ |
Additional Notes
- The integrand is defined for all real $x$.
- This follows the same reduce-the-power-of-$x$ pattern as Problem 165, but requires only two rounds of integration by parts instead of four since the power starts at $x^2$ rather than $x^4$.
- The result can be checked by differentiating $-\tfrac12x^2\cos2x+\tfrac12x\sin2x+\tfrac14\cos2x$ and confirming it returns $x^2\sin2x$.
Result
The solution to the integral is: