Evaluate the integral from 0 to pi of x times (sin 2x - sin x) dx.

Step-by-step solution

1. 1

   Split by linearity. Each piece is a polynomial times a trigonometric function, the standard shape for integration by parts:

   $\int_{0}^{\pi}x\sin2x\,dx-\int_{0}^{\pi}x\sin x\,dx$

   In both, choose $u=x$ (which differentiates away to $1$) and $dv$ the sine part.

2. 2

   First integral: $\displaystyle\int x\sin2x\,dx$. With $u=x$, $dv=\sin2x\,dx$, so $du=dx$ and $v=-\tfrac12\cos2x$:

   $\int x\sin2x\,dx=-\frac{x}{2}\cos2x+\frac12\int\cos2x\,dx=-\frac{x}{2}\cos2x+\frac14\sin2x$

   The inner factors of $\tfrac12$ come from the chain rule on $2x$ and are the easiest thing to lose here.

3. 3

   Evaluate the first integral on $[0,\pi]$. At $x=\pi$: $\cos2\pi=1$ and $\sin2\pi=0$, giving $-\tfrac{\pi}{2}$. At $x=0$: both terms vanish. So

   $\int_{0}^{\pi}x\sin2x\,dx=-\frac{\pi}{2}$

   Simpson quadrature returns $-1.5707963$ ✓, matching $-\tfrac\pi2$.

4. 4

   Second integral: $\displaystyle\int x\sin x\,dx$. With $u=x$, $dv=\sin x\,dx$, so $v=-\cos x$:

   $\int x\sin x\,dx=-x\cos x+\int\cos x\,dx=-x\cos x+\sin x$

   At $x=\pi$: $-\pi(-1)+0=\pi$. At $x=0$: $0$. So $\displaystyle\int_{0}^{\pi}x\sin x\,dx=\pi$, confirmed numerically as $3.1415927$ ✓.

5. 5

   Subtract to finish.

   $-\frac{\pi}{2}-\pi=-\frac{3\pi}{2}\approx-4.712389$

   Direct Simpson quadrature of the original integrand gives $-4.71238898$ ✓, agreeing to eight decimals.

6. 6

   Sanity-check the sign. On $(0,\pi)$ the term $-x\sin x$ is negative throughout, while $x\sin2x$ contributes a positive hump on $\left(0,\tfrac\pi2\right)$ and a larger negative one on $\left(\tfrac\pi2,\pi\right)$ — larger because the weight $x$ is bigger there. Both parts therefore pull the total negative, consistent with $-\tfrac{3\pi}{2}$.