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How do I integrate with substitution only this integrand? :

$$ \int \frac{\sin\sqrt x}{\sqrt x} \ dx$$

I tried to solve it with the fact that $$(\sqrt x )'= \frac {1} {2 \sqrt x}$$ But got lost.

Thanks for your help.

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    $\begingroup$ what is "sen" ? $\endgroup$ – R.N Sep 17 '15 at 20:26
  • $\begingroup$ @RaziehNoori It's commonly used in Italy for the sine function, that we call “seno”. I hope this usage will vanish. $\endgroup$ – egreg Sep 17 '15 at 20:43
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You may just substitute $\sqrt{x}=u$, $\dfrac{1}{2\sqrt{x}}dx=du$ giving $$ \int\frac{\sin (\sqrt{x})}{\sqrt{x}}dx=2\int\sin (u)\:du=-2\cos (u)+C=-2\cos (\sqrt{x})+C. $$

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Let $f(x)=-\cos(x)$ and $g(x)=\sqrt{x}$. Then you may write your intégral as $$ \int \frac{2}{2}\cdot\frac{\sin(\sqrt{x})}{\sqrt{x}}dx=\int 2f^{'}(g(x))g^{'}(x). $$

From the chain rule, we know an antiderivative for $f^{'}(g(x))g^{'}(x)$ given by $f(g(x))$. Your integral is then $$ \int \cdot\frac{\sin(\sqrt{x})}{\sqrt{x}}dx=-2(f(g(x))+C)=-2\cos(\sqrt{x})+C $$ Note that the rightmost $C$ is different than the middle one, but it doesn't matter as $2C$ can be seen as an integration constant.

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Notice, let $$\sqrt x=t\implies \frac{dx}{2\sqrt x}=dt$$$$\frac{dx}{2t}=dt\iff dx=2tdt$$ Now, substituting $\sqrt x=t$ & $dx=2tdt$ we get $$\int \frac{\sin(\sqrt x) \ dx}{\sqrt x}dx$$$$=\int \frac{\sin t\ (2tdt)}{t}$$ $$=2\int \sin t\ dt=2(-\cos t)+C$$ $$=\color{red}{-2\cos (\sqrt x)+C}$$

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