Step 1: Understanding the Question:
We need to find the indefinite integral of the product of an algebraic function $x^2$ and a trigonometric function $\cos x$. Step 2: Key Formula or Approach:
Use Integration by Parts (IBP) rule: $\int u v dx = u \int v dx - \int (u' \int v dx) dx$.
According to LIATE rule, $u = x^2$ (algebraic) and $v = \cos x$ (trigonometric). Step 3: Detailed Explanation:
Let $I = \int x^2 \cos x dx$.
Apply IBP: $u = x^2 \implies du = 2x dx$; $dv = \cos x dx \implies v = \sin x$.
\[ I = x^2 \sin x - \int (2x) \sin x dx \]
For the second integral $\int 2x \sin x dx$, apply IBP again:
$u = 2x \implies du = 2 dx$; $dv = \sin x dx \implies v = -\cos x$.
\[ \int 2x \sin x dx = (2x)(-\cos x) - \int (2)(-\cos x) dx \]
\[ = -2x \cos x + 2 \int \cos x dx = -2x \cos x + 2 \sin x \]
Substituting this back into the first expression:
\[ I = x^2 \sin x - (-2x \cos x + 2 \sin x) + c \]
\[ I = x^2 \sin x + 2x \cos x - 2 \sin x + c \]
Step 4: Final Answer:
The integral is $x^2 \sin x + 2x \cos x - 2 \sin x + c$.