1. The problem is to find the integral $\int t^2 \cos t \, dt$. 2. We use integration by parts, which states: $$\int u \, dv = uv - \int v \, du$$ 3. Choose $u = t^2$ and $dv = \cos t \, dt$. 4. Then, $du = 2t \, dt$ and $v = \sin t$. 5. Apply integration by parts: $$\int t^2 \cos t \, dt = t^2 \sin t - \int 2t \sin t \, dt$$ 6. Now, we need to evaluate $\int 2t \sin t \, dt$. 7. Use integration by parts again for $\int 2t \sin t \, dt$: - Let $u = 2t$, so $du = 2 dt$. - Let $dv = \sin t \, dt$, so $v = -\cos t$. 8. Then: $$\int 2t \sin t \, dt = -2t \cos t + \int 2 \cos t \, dt$$ 9. Evaluate $\int 2 \cos t \, dt = 2 \sin t$. 10. Substitute back: $$\int 2t \sin t \, dt = -2t \cos t + 2 \sin t$$ 11. Substitute into the original integral: $$\int t^2 \cos t \, dt = t^2 \sin t - (-2t \cos t + 2 \sin t) + C$$ 12. Simplify: $$= t^2 \sin t + 2t \cos t - 2 \sin t + C$$ Final answer: $$\int t^2 \cos t \, dt = t^2 \sin t + 2t \cos t - 2 \sin t + C$$