Question

\[ \int x^2 \cos x d x= \]

• A. \(x^{2}\mathrm{sin}x+2x\mathrm{cos}x-2\mathrm{sin}x+\mathrm{c}\), where c is the constant of integration
• B. \(x^{2}\mathrm{sin}x-2x\mathrm{cos}x-2\mathrm{sin}x+\mathrm{c}\), where c is the constant of integration
• C. \(x^{2}\mathrm{sin}x-2x\mathrm{cos}x+2\mathrm{sin}x+\mathrm{c}\), where c is the constant of integration
• D. \(x^{2}\mathrm{sin}x+2x\mathrm{cos}x+2\mathrm{sin}x+\mathrm{c}\), where \(c\) is the constant of integration

Hint:

Apply integration by parts twice, lowering the power of x each time.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Use \(\int u\,dv=uv-\int v\,du\) with \(u=x^{2}\) and \(dv=\cos x\,dx\).

Step 2: First round
\(\int x^{2}\cos x\,dx=x^{2}\sin x-\int2x\sin x\,dx\).

Step 3: Second round
\(\int x\sin x\,dx=-x\cos x+\int\cos x\,dx=-x\cos x+\sin x\).

Step 4: Combine
\(x^{2}\sin x-2(-x\cos x+\sin x)+c=x^{2}\sin x+2x\cos x-2\sin x+c\).

Final Answer:
The integral is x^2 sin x + 2x cos x - 2 sin x + c. \[ \boxed{\text{(A) }x^{2}\sin x+2x\cos x-2\sin x+\mathrm{c}} \]