Question

The value of \(\int 2x^{\frac{1}{3}}sin\sqrt[3]{x^2}\,dx\) is

• A. \((x^{\frac{2}{3}}cosx^{\frac{2}{3}}+sinx^{\frac{2}{3}})+c\)
• B. \(3[x^{\frac{2}{3}}cosx^{\frac{2}{3}}-sinx^{\frac{2}{3}}]+c\)
• C. \(3[-x^{\frac{2}{3}}cosx^{\frac{2}{3}}-sinx^{\frac{2}{3}}]+c\)
• D. \(3[-x^{\frac{2}{3}}cosx^{\frac{2}{3}}+sinx^{\frac{2}{3}}]+c\)

Hint:

Put \(t=x^{2/3}\) and integrate by parts.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Let \(t=x^{2/3}\). Then \(dt=\tfrac23x^{-1/3}dx\), so \(dx=\tfrac32x^{1/3}dt\).

Step 2: Key Formula or Approach
The integrand becomes \(2x^{1/3}\sin t\cdot\tfrac32x^{1/3}dt=3x^{2/3}\sin t\,dt=3t\sin t\,dt\).

Step 3: Detailed Explanation
By parts: \(\int t\sin t\,dt=-t\cos t+\sin t\).
\[ \int2x^{1/3}\sin x^{2/3}\,dx=3\left[-t\cos t+\sin t\right]+c \]
\[ =3\left[-x^{2/3}\cos x^{2/3}+\sin x^{2/3}\right]+c \]

Final Answer:
The integral is \(3[-x^{2/3}\cos x^{2/3}+\sin x^{2/3}]+c\), option (D). \[ \boxed{3\left[-x^{2/3}\cos x^{2/3}+\sin x^{2/3}\right]+c\ \text{(D)}} \]