Find the indefinite integral of x^2 * sin(x) with respect to x.

Step-by-step solution

1. 1

   Choose $u$ so that differentiating reduces the polynomial. In $\int u\,dv=uv-\int v\,du$ the aim is a simpler leftover integral. Taking $u=x^{2}$ makes $du=2x\,dx$, dropping the degree from $2$ to $1$; taking $u=\sin x$ instead would raise the polynomial degree and make things worse. This is the LIATE guideline in action — the algebraic factor is the one to differentiate.

2. 2

   First application. With $u=x^{2}$, $dv=\sin x\,dx$ (so $v=-\cos x$):

   $\int x^{2}\sin x\,dx=-x^{2}\cos x-\int\left(-\cos x\right)(2x)\,dx=-x^{2}\cos x+\int 2x\cos x\,dx.$

   The two minus signs — one from $v=-\cos x$ and one from the formula — combine to a plus, which is where sign errors usually creep in.

3. 3

   Second application on the remaining integral. For $\int x\cos x\,dx$ take $u=x$, $dv=\cos x\,dx$, so $du=dx$, $v=\sin x$:

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

   The polynomial degree is now $0$, so the process terminates — a degree-$n$ polynomial needs exactly $n$ rounds of parts.

4. 4

   Combine the pieces. Multiplying the second result by the factor $2$ and substituting:

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

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

5. 5

   Verify by differentiating. Using the product rule on each term:

   $\frac{d}{dx}\left(-x^{2}\cos x\right)=-2x\cos x+x^{2}\sin x,\quad \frac{d}{dx}\left(2x\sin x\right)=2\sin x+2x\cos x,\quad \frac{d}{dx}\left(2\cos x\right)=-2\sin x.$

   Adding: the $\mp2x\cos x$ pair cancels, the $\pm2\sin x$ pair cancels, leaving exactly $x^{2}\sin x$ ✓. A numeric check at $x=1.3$ gives $1.6284133$ both ways ✓.