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3.3 · Q127

Q.Evaluate: ∫cos⁡(x3) dx\int \cos(\sqrt[3]{x})\,dx

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Let t=x1/3t=x^{1/3}, so x=t3x=t^3, dx=3t2dtdx=3t^2dt:

∫cos⁡(x1/3)dx=∫cos⁡t⋅3t2dt=3∫t2cos⁡t dt\int\cos\left(x^{1/3}\right)dx=\int\cos t\cdot3t^2dt=3\int t^2\cos t\,dt

By parts (u=t2u=t^2, dv=cos⁡t dt⇒v=sin⁡tdv=\cos t\,dt\Rightarrow v=\sin t): ∫t2cos⁡t dt=t2sin⁡t−2∫tsin⁡t dt\int t^2\cos t\,dt=t^2\sin t-2\int t\sin t\,dt.

By parts again (u=tu=t, dv=sin⁡t dt⇒v=−cos⁡tdv=\sin t\,dt\Rightarrow v=-\cos t): ∫tsin⁡t dt=−tcos⁡t+sin⁡t\int t\sin t\,dt=-t\cos t+\sin t.

So ∫t2cos⁡t dt=t2sin⁡t−2(−tcos⁡t+sin⁡t)=t2sin⁡t+2tcos⁡t−2sin⁡t\int t^2\cos t\,dt=t^2\sin t-2(-t\cos t+\sin t)=t^2\sin t+2t\cos t-2\sin t. …

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