Question

The value of \(\displaystyle\int x\cos x\,dx\) is:

• A. \(\cos x+x\sin x+c\)
• B. \(x\sin x+c\)
• C. \(x\cos x+c\)
• D. \(\sin x+x\cos x+c\)

Hint:

Integrate by parts with \(u=x\), \(dv=\cos x\,dx\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Use integration by parts, \(\int u\,dv=uv-\int v\,du\), with \(u=x\) (algebraic, differentiate) and \(dv=\cos x\,dx\) (trigonometric, integrate) — ILATE order.

Step 2: Applying integration by parts:
\(du=dx\), \(v=\sin x\). So \(\int x\cos x\,dx=x\sin x-\int \sin x\,dx\).

Step 3: Finishing the remaining integral:
\(\int \sin x\,dx=-\cos x\). So the result is \(x\sin x-(-\cos x)+c=x\sin x+\cos x+c\).

Final Answer:
\(\displaystyle\int x\cos x\,dx=\boxed{\cos x+x\sin x+c}\).