Integrate \(3x^2 (cosx^3+8)\).

(a) \(sinx^3-8x^3+C\)

(b) \(sinx^3+8x^3+C\)

(c) –\(sinx^3+8x^3+C\)

(d) \(sinx^3-x^3+C\)

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The correct answer is (b) \(sinx^3+8x^3+C\)

Easy explanation: By using the method of integration by substitution,

Let x^3=t

Differentiating w.r.t x, we get

3x^2 dx=dt

\(\int 3x^2 \,(cosx^3+8) \,dx=\int (cost+8)dt\)

\(\int (cost+8) dt=sint+8t\)

Replacing t with x^3,we get

\(\int 3x^2 (cosx^3+8) dx=sinx^3+8x^3+C\)