8. \( \int x^{2} \cos x d x \)

To solve the integral \( \int x^{2} \cos x \, dx \), we can use integration by parts. Let \( u = x^2 \) and \( dv = \cos x \, dx \). Thus, \( du = 2x \, dx \) and \( v = \sin x \). Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \): \[ \int x^{2} \cos x \, dx = x^{2} \sin x - \int 2x \sin x \, dx \] Now we need to integrate \( \int 2x \sin x \, dx \), using integration by parts again. Let \( u = 2x \) and \( dv = \sin x \, dx \). Then, \( du = 2 \, dx \) and \( v = -\cos x \). So we have: \[ \int 2x \sin x \, dx = -2x \cos x - \int -2 \cos x \, dx = -2x \cos x + 2 \sin x \] Putting this back, we get: \[ \int x^{2} \cos x \, dx = x^{2} \sin x + 2x \cos x - 2 \sin x + C \] where \( C \) is the constant of integration. Therefore, the final result is: \[ \int x^{2} \cos x \, dx = x^{2} \sin x + 2x \cos x - 2 \sin x + C \]