∫ x^2cos≤ft(x^3)dx is equal to (where C is a constant of integration)

The correct answer is Option (4) → \(\displaystyle\frac{1}{3}\sin\left(x^3\right) + C\)

The factor \(x^2\) is, apart from a constant, the derivative of \(x^3\), so the substitution \(t = x^3\) works.

    \(dt = 3x^2\,dx\)
    \(x^2\,dx = \frac{1}{3}\,dt\)

So the integral becomes

    \( \displaystyle \int \cos t \cdot \frac{1}{3}\,dt = \frac{1}{3}\sin t + C \)

    \( \displaystyle = \frac{1}{3}\sin\left(x^3\right) + C \)

Check: the derivative of \(\frac{1}{3}\sin\left(x^3\right)\) is \(\frac{1}{3}\cos\left(x^3\right) \cdot 3x^2 = x^2\cos\left(x^3\right)\).