The value of 2 x¹/3} [3]{x²} d x is

Correct Answer is : 3[−x23cos⁡x23+sin⁡x23]+c3≤ft[-x²/3} x²/3}+ x²/3}]+c3[−x32cosx32+sinx32]+c

3≤ft[x²/3} x²/3}- x²/3}]+c

3≤ft[-x²/3} x²/3}- x²/3}]+c

3≤ft[-x²/3} x²/3}+ x²/3}]+c

Correct Answer is : 3[−x23cos⁡x23+sin⁡x23]+c3≤ft[-x²/3} x²/3}+ x²/3}]+c3[−x32cosx32+sinx32]+c

Test Index

MHTCET 17 Apr 2026 Shift 1 Paper

Question : 130 of 150

Marks: +1, -0

The value of

\int 2 x^{\frac{1}{3}} \sin \sqrt[3]{x^{2}} d x

Solution:

Concept:

Substitute

t = x^{2/3}

to convert the integral into a simpler form, then use integration by parts.

Explanation:

Let

I = \int 2x^{1/3} \sin\sqrt[3]{x^2}\,dx

Since

\sqrt[3]{x^2} = x^{2/3}

, put

t = x^{2/3}

Differentiate:

dt = \frac{2}{3}x^{-1/3}dx

So,

dx = \frac{3}{2}x^{1/3}dt

Substitute into the integral:

I = \int 2x^{1/3}\sin t \cdot \frac{3}{2}x^{1/3}dt

This simplifies to

I = 3\int x^{2/3}\sin t\,dt = 3\int t\sin t\,dt

Using integration by parts:

\int t\sin t\,dt = -t\cos t + \sin t

Therefore,

I = 3\left[-t\cos t + \sin t\right] + c

Substitute

t = x^{2/3}

back:

I = 3\left[-x^{2/3}\cos x^{2/3} + \sin x^{2/3}\right] + c

Answer:

Option D:

3\left[-x^{\frac{2}{3}} \cos x^{\frac{2}{3}}+\sin x^{\frac{2}{3}}\right]+c

Go to Question:

Prev Question

Next Question