Question

$\int x^{2}\cos x~dx=$

• A. $x^{2}\sin x+2x \cos x-2 \sin x+c$
• B. $x^{2}\sin x-2x \cos x-2 \sin x+c$
• C. $x^{2}\sin x-2x \cos x+2 \sin x+c$
• D. $x^{2}\sin x+2x \cos x+2 \sin x+c$

Hint:

The sign pattern for $x^n \cos x$ type integration follows the tabular method (LIATE).

Answer 1

The Correct Option is A

Step 1: Concept
Use the Integration by Parts rule: $\int u v~dx = u \int v~dx - \int (u' \int v~dx) dx$.

Step 2: Analysis
Let $u = x^2$ and $dv = \cos x~dx$.
$\int x^2 \cos x~dx = x^2 \sin x - \int 2x \sin x~dx$.

Step 3: Calculation
Apply parts again to $\int 2x \sin x~dx$:
$\int 2x \sin x~dx = 2x(-\cos x) - \int 2(-\cos x) dx = -2x \cos x + 2 \sin x$.
Substituting back: $x^2 \sin x - (-2x \cos x + 2 \sin x) = x^2 \sin x + 2x \cos x - 2 \sin x + c$.

Step 4: Conclusion
Hence, the correct expression is $x^2 \sin x + 2x \cos x - 2 \sin x + c$. Final Answer: (A)