Question:
The value of integral \(\int x^3cosx\,dx\) is...
The value of integral \(\int x^3cosx\,dx\) is...
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Apply integration by parts repeatedly, lowering the power of x each time.
Updated On: Oct 1, 2026
- \(x^3sinx+x^2cosx-6xsinx+6cosx+c\)
- \(x^3sinx+3x^2sinx-6xsinx-6cosx+c\)
- \(x^3sinx+3x^2cosx-6xsinx-6cosx+c\)
- \(x^3sinx+3x^2cosx-6xsinx+6cosx+c\)
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The Correct Option is C
Solution and Explanation
Step 1: Understand the concept
We use \(\int u\,dv = uv - \int v\,du\) and take \(u = x^n\) each time, so the power of \(x\) falls by one at every step.
Step 2: First two steps
\[ \int x^3\cos x\,dx = x^3\sin x - 3\int x^2\sin x\,dx \]
\[ \int x^2\sin x\,dx = -x^2\cos x + 2\int x\cos x\,dx \]
Step 3: Next step
\(\int x\cos x\,dx = x\sin x + \cos x\). Substitute back: \(\int x^2\sin x\,dx = -x^2\cos x + 2x\sin x + 2\cos x\).
Step 4: Combine
\[ x^3\sin x - 3\left(-x^2\cos x + 2x\sin x + 2\cos x\right) = x^3\sin x + 3x^2\cos x - 6x\sin x - 6\cos x + c \]
This is option (C). Differentiating it gives back \(x^3\cos x\), since the lower terms cancel in pairs.
Final Answer:
The integral is x^3 sin x + 3x^2 cos x - 6x sin x - 6 cos x + c. This is option (C). \[ \boxed{\text{(C) }x^3\sin x+3x^2\cos x-6x\sin x-6\cos x+c} \]
We use \(\int u\,dv = uv - \int v\,du\) and take \(u = x^n\) each time, so the power of \(x\) falls by one at every step.
Step 2: First two steps
\[ \int x^3\cos x\,dx = x^3\sin x - 3\int x^2\sin x\,dx \]
\[ \int x^2\sin x\,dx = -x^2\cos x + 2\int x\cos x\,dx \]
Step 3: Next step
\(\int x\cos x\,dx = x\sin x + \cos x\). Substitute back: \(\int x^2\sin x\,dx = -x^2\cos x + 2x\sin x + 2\cos x\).
Step 4: Combine
\[ x^3\sin x - 3\left(-x^2\cos x + 2x\sin x + 2\cos x\right) = x^3\sin x + 3x^2\cos x - 6x\sin x - 6\cos x + c \]
This is option (C). Differentiating it gives back \(x^3\cos x\), since the lower terms cancel in pairs.
Final Answer:
The integral is x^3 sin x + 3x^2 cos x - 6x sin x - 6 cos x + c. This is option (C). \[ \boxed{\text{(C) }x^3\sin x+3x^2\cos x-6x\sin x-6\cos x+c} \]
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