To solve \(\int_{0}^{\pi} x^2 \sin x \, dx\), we use integration by parts. Let \(u = x^2\) and \(dv = \sin x \, dx\). Then \(du = 2x \, dx\) and \(v = -\cos x\).
Applying integration by parts:
\(\int u \, dv = uv - \int v \, du\)
\(\int x^2 \sin x \, dx = -x^2 \cos x + \int 2x \cos x \, dx\)
Now, integrate \(\int 2x \cos x \, dx\) by parts again. Let \(u = 2x\) and \(dv = \cos x \, dx\). Then \(du = 2 \, dx\) and \(v = \sin x\).
\(\int 2x \cos x \, dx = 2x \sin x - \int 2 \sin x \, dx\)
\(= 2x \sin x + 2 \cos x\)
Substituting back, we have:
\(\int x^2 \sin x \, dx = -x^2 \cos x + 2x \sin x + 2 \cos x\)
Evaluate from 0 to \(\pi\):
\(\left[ -x^2 \cos x + 2x \sin x + 2 \cos x \right]_{0}^{\pi}\)
At \(x = \pi\):
\(-\pi^2 \cos \pi + 2\pi \sin \pi + 2 \cos \pi = \pi^2 + 2(-1) = \pi^2 - 2\)
At \(x = 0\):
\(-0^2 \cos 0 + 2 \cdot 0 \cdot \sin 0 + 2 \cos 0 = 2\)
Subtracting, we get:
\((\pi^2 - 2) - 2 = \pi^2 - 4\)
