∫sin⁡(x3)dx=a2−x2/3cos⁡(x3)+3b x3⋅sin⁡(x3)+c

I=∫sin⁡(x3)dx put x3=t⇒13x-23dx=dt I=3∫t2sin⁡(t) dt =3-t2cost+2tsint+2cost+c =32-t2cost+6tsint+c =32-x23cosx13+6x13sinx13+c

∫sin⁡(x3)dx=a2−x2/3cos⁡(x3)+3b x3⋅sin⁡(x3)+c

I=∫sin⁡(x3)dx put x3=t⇒13x-23dx=dt I=3∫t2sin⁡(t) dt =3-t2cost+2tsint+2cost+c =32-t2cost+6tsint+c =32-x23cosx13+6x13sinx13+c

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$\int \sin (\sqrt[3]{x}) dx = a \{(2 - x^{2 / 3}) \cos (\sqrt[3]{x}) + 3 b^{} \sqrt[3]{x} \cdot \sin (\sqrt[3]{x})\} + c$

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a = 1

a = 3

b= 2

b = 1

answer is A, B.

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Detailed Solution

$I = \int \sin (\sqrt[3]{x}) dx p u t \sqrt[3]{x} = t \Rightarrow \frac{1}{3} x^{- \frac{2}{3}} d x = d t I = 3 \int t^{2} \sin (t) d t = 3 [- t^{2} \cos t + 2 t \sin t + 2 \cos t] + c = 3 (2 - t^{2}) \cos t + 6 t \sin t + c = 3 (2 - x^{\frac{2}{3}}) \cos x^{\frac{1}{3}} + 6 x^{\frac{1}{3}} \sin x^{\frac{1}{3}} + c$