The value of ∫sin⁡x3dx is

Put x1/3=θ so that dx=3θ2dθ∴I=3∫θ2sin⁡θdθ=−3θ2cos⁡θ+6∫θcos⁡θdθ=−3θ2cos⁡θ+6θsin⁡θ−6∫sin⁡θdθ=32−θ2cos⁡θ+6θsin⁡θ+C.

The value of ∫sin⁡x3dx is

Put x1/3=θ so that dx=3θ2dθ∴I=3∫θ2sin⁡θdθ=−3θ2cos⁡θ+6∫θcos⁡θdθ=−3θ2cos⁡θ+6θsin⁡θ−6∫sin⁡θdθ=32−θ2cos⁡θ+6θsin⁡θ+C.

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The value of $\int \sin \sqrt[3]{x} d x$ is

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$3 \{(2 - x^{2 / 3}) \cos x^{1 / 3} + 2 x^{1 / 3} \sin x^{1 / 3}\} + C$

$\{(2 - x^{2 / 3}) \cos x^{1 / 3} + 2 x^{1 / 3} \sin x^{1 / 3}\} + C$

none of these

$3 \{(2 - x^{2 / 3}) \sin x^{1 / 3} - 2 x^{1 / 3} \cos x^{1 / 3}\} + C$

answer is B.

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Detailed Solution

Put $x^{1 / 3} = θ$ so that $d x = 3 θ^{2} d θ$

$\begin{array}{l} ∴ I = 3 \int θ^{2} \sin θ d θ \\ = - 3 θ^{2} \cos θ + 6 \int θ \cos θ d θ \\ = - 3 θ^{2} \cos θ + 6 θ \sin θ - 6 \int \sin θ d θ \\ = 3 (2 - θ^{2}) \cos θ + 6 θ \sin θ + C . \end{array}$

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The value of ∫sin⁡x3dx is

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The value of $\int \sin \sqrt[3]{x} d x$ is