The value of int 2x^{1/3} sin x^{2/3} dx is | vedclass

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(D) Let $I = \int 2x^{1/3} \sin x^{2/3} dx$.Substitute $u = x^{2/3}$, then $du = \frac{2}{3} x^{-1/3} dx$, which implies $dx = \frac{3}{2} x^{1/3} du$

Vedclass · Accepted Answer

$3[-x^{2/3} \cos x^{2/3} + \sin x^{2/3}] + c$

Vedclass · Answer

(D) Let $I = \int 2x^{1/3} \sin x^{2/3} dx$.Substitute $u = x^{2/3}$, then $du = \frac{2}{3} x^{-1/3} dx$, which implies $dx = \frac{3}{2} x^{1/3} du$.Substituting into the integral: $I = \int 2x^{1/3} \sin(u) \cdot \frac{3}{2} x^{1/3} du = 3 \int x^{2/3} \sin(u) du = 3 \int u \sin u du$.Using integration by parts $\int u \sin u du = u(-\cos u) - \int 1 \cdot (-\cos u) du = -u \cos u + \sin u$.Thus, $I = 3[-u \cos u + \sin u] + c$.Substituting $u = x^{2/3}$ back, $I = 3[-x^{2/3} \cos x^{2/3} + \sin x^{2/3}] + c$.

The value of $\int 2x^{1/3} \sin x^{2/3} dx$ is

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