From Förberedande kurs i matematik 2

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-	{{NAVCONTENT_START}}	+	Let's rewrite the integral somewhat,
-	<center> [[Image:2_2_3f.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>2\sin\sqrt{x}\cdot\frac{1}{2\sqrt{x}}\,\textrm{.}</math>}}
		+
		+	Here, we see that the factor on the right, <math>1/2\sqrt{x}</math>, is the derivative of the expression <math>\sqrt{x}</math>, which appears in the factor on the left, <math>2\sin \sqrt{x}\,</math>. With the substitution <math>u=\sqrt{x}</math>, the integrand can therefore be written as
		+
		+	{{Displayed math||<math>2\sin u\cdot u'</math>}}
		+
		+	and the integral becomes
		+
		+	{{Displayed math||<math>\begin{align}
		+	\int \frac{\sin \sqrt{x}}{\sqrt{x}}\,dx
		+	&= \left\{ \begin{align}
		+	u &= \sqrt{x}\\[5pt]
		+	du &= (\sqrt{x}\,)'\,dx = \frac{1}{2\sqrt{x}}\,dx
		+	\end{align}\, \right\}\\[5pt]
		+	&= 2\int \sin u\,du\\[5pt]
		+	&= -2\cos u+C\\[5pt]
		+	&= -2\cos\sqrt{x} + C\,\textrm{.}
		+	\end{align}</math>}}

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Current revision

Let's rewrite the integral somewhat,

\displaystyle 2\sin\sqrt{x}\cdot\frac{1}{2\sqrt{x}}\,\textrm{.}

Here, we see that the factor on the right, \displaystyle 1/2\sqrt{x}, is the derivative of the expression \displaystyle \sqrt{x}, which appears in the factor on the left, \displaystyle 2\sin \sqrt{x}\,. With the substitution \displaystyle u=\sqrt{x}, the integrand can therefore be written as

\displaystyle 2\sin u\cdot u'

and the integral becomes

\displaystyle \begin{align} \int \frac{\sin \sqrt{x}}{\sqrt{x}}\,dx &= \left\{ \begin{align} u &= \sqrt{x}\\[5pt] du &= (\sqrt{x}\,)'\,dx = \frac{1}{2\sqrt{x}}\,dx \end{align}\, \right\}\\[5pt] &= 2\int \sin u\,du\\[5pt] &= -2\cos u+C\\[5pt] &= -2\cos\sqrt{x} + C\,\textrm{.} \end{align}