Evaluate the integral of x^2 * sin(x) from x = 0 to x = pi.

Step-by-step solution

1. 1

   Choose $u$ so the polynomial gets smaller. With a polynomial times a trigonometric function, always let $u$ be the polynomial: differentiating lowers its degree while $\sin x$ integrates to $-\cos x$ without growing. Take

   $u=x^{2},\quad dv=\sin x\,dx\ \Longrightarrow\ du=2x\,dx,\quad v=-\cos x$

   Choosing the other way round would raise the degree and never terminate.

2. 2

   Apply integration by parts once.

   $\int x^{2}\sin x\,dx=-x^{2}\cos x+\int 2x\cos x\,dx$

   The remaining integral has degree $1$ instead of $2$, so one more pass will finish it.

3. 3

   Apply it a second time to $\int x\cos x\,dx$. Now $u=x$, $dv=\cos x\,dx$, so $du=dx$ and $v=\sin x$:

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

   Hence $\int 2x\cos x\,dx=2x\sin x+2\cos x$.

4. 4

   Write the full antiderivative.

   $\int x^{2}\sin x\,dx=-x^{2}\cos x+2x\sin x+2\cos x+C$

   (Degree 2 needed exactly two passes; a degree-$n$ polynomial needs $n$.)

5. 5

   Substitute the limits. At $x=\pi$, using $\cos\pi=-1$ and $\sin\pi=0$:

   $-\pi^{2}(-1)+0+2(-1)=\pi^{2}-2$

   At $x=0$, using $\cos 0=1$ and $\sin 0=0$:

   $0+0+2=2$

6. 6

   Subtract and check.

   $\int_{0}^{\pi}x^{2}\sin x\,dx=\left(\pi^{2}-2\right)-2=\pi^{2}-4\approx 5.8696$

   A numerical Simpson estimate gives $5.8696044011$, matching $\pi^{2}-4=5.8696044011$ ✓. The answer is positive, as it must be: $\sin x\ge 0$ throughout $[0,\pi]$.