What is the integral of cos cube root x dx??

Experts, I'm getting to you for the 2nd time.......plzz Reply!!

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{I}=\int \cos {\mathrm{x}}^{\frac{1}{3}}\mathrm{dx} \\ \mathrm{Let} \mathrm{x}={\mathrm{u}}^3 \\ \Rightarrow \mathrm{dx}=3{\mathrm{u}}^2\mathrm{du} \\ \Rightarrow \mathrm{I}=3\int \mathrm{sinu}.{\mathrm{u}}^2\mathrm{du} \\ \mathrm{Using} \mathrm{integration} \mathrm{by} \mathrm{parts} \mathrm{we} \mathrm{get}, \\ \mathrm{I}=3\left[{\mathrm{u}}^2\int \mathrm{sinu}.\mathrm{du}-\int \left\{\frac{\mathrm{d}}{\mathrm{du}}\left({\mathrm{u}}^2\right)\int \mathrm{sinu}.\mathrm{du}\right\}\mathrm{du}\right] \\ =3\left[-{\mathrm{u}}^2.\mathrm{cosu}+\int \left\{2\mathrm{u}.\mathrm{cosudu}\right\}\right] \\ =3\left[-{\mathrm{u}}^2.\mathrm{cosu}+2\int \mathrm{u}.\mathrm{cosudu}\right] \\ \mathrm{Again} \mathrm{using} \mathrm{integration} \mathrm{by} \mathrm{parts} \mathrm{we} \mathrm{get}, \\ \mathrm{I}=3\left[-{\mathrm{u}}^2.\mathrm{cosu}+2\left\{\mathrm{u}.\int \mathrm{cosu}.\mathrm{du}-\int \left(\frac{\mathrm{d}}{\mathrm{du}}.\mathrm{u}\int \mathrm{cosu}.\mathrm{du}\right)\mathrm{du}\right\}\right] \\ =3\left[-{\mathrm{u}}^2.\mathrm{cosu}+2\left\{\mathrm{u}.\mathrm{sinu}-\int \mathrm{sinu}.\mathrm{du}\right\}\right] \\ =3\left[-{\mathrm{u}}^2.\mathrm{cosu}+2\left\{\mathrm{u}.\mathrm{sinu}+\mathrm{cosu}\right\}\right]+\mathrm{C} \\ \mathrm{I}=-3{\mathrm{u}}^2.\mathrm{cosu}+6\mathrm{usinu}+6\mathrm{cosu}+\mathrm{C} \\ \mathrm{As} \mathrm{x}={\mathrm{u}}^3 \\ \mathbf{\Rightarrow }\mathbf{\int }\mathbf{cos}\mathbf{ }{\mathbf{x}}^{\frac{\mathbf{1}}{\mathbf{3}}}\mathbf{dx}\mathbf{=}\mathbf{-}\mathbf{3}{\mathbf{x}}^{\frac{\mathbf{2}}{\mathbf{3}}}\mathbf{.}{\mathbf{cosx}}^{\frac{\mathbf{1}}{\mathbf{3}}}\mathbf{+}\mathbf{6}{\mathbf{x}}^{\frac{\mathbf{1}}{\mathbf{3}}}\mathbf{.}{\mathbf{sinx}}^{\frac{\mathbf{1}}{\mathbf{3}}}\mathbf{+}\mathbf{6}{\mathbf{cosx}}^{\frac{\mathbf{1}}{\mathbf{3}}}\mathbf{+}\mathbf{C} \end{gathered}$$

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