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I asked my professor for the integration of $\sqrt{x} \sin x$ but he said that this is not in our syllabus. I tried on my own by substitution and by parts but it gets more complicated.

Please help me

Thanks!

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  • $\begingroup$ Why! That's not closed-form integrable! $\endgroup$ Jul 17, 2017 at 7:18
  • $\begingroup$ try typing integrate $\sqrt{x} \sin x$ in wolframalpha.com $\endgroup$ Jul 17, 2017 at 7:36

2 Answers 2

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By parts,

$$\int\sqrt x\sin x\,dx=\sqrt x\cos x-\int\frac{\cos x}{2\sqrt x}dx.$$

Then by the substitution $x=t^2$,

$$\int\frac{\cos x}{2\sqrt x}dx=\int (\cos (t^2)\,)dt.$$

This is a well-known Fresnel integral, for which no closed-form expression exists (this can be formally proven using Liouville's theorem).

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As said, this is slightly more difficult than what you learnt up to now.

You want to compute $$I=\int \sqrt x \sin(x)\,dx$$ First, let $$x=y^2 \implies dx=2y dy$$ which makes $$I=2\int y^2\sin(y^2)\,dy$$ Use one integration by parts $$u'=y \sin(y^2)\,dy\implies u=-\frac 12\cos(y^2)$$ $$v=y\implies v'=dy$$ which make $$I=2\left(-\frac 12y\cos(y^2)+\frac 12\int \cos(y^2)\,dy\right)=-y\cos(y^2)+\int \cos(y^2)\,dy$$ and here appears the Fresnel integral which is a special function.

The final result would then be $$I=\sqrt{\frac{\pi }{2}} C\left(\sqrt{\frac{2}{\pi }} y\right)-y \cos \left(y^2\right)$$

Don't worry : you will learn about them sooner or later !

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