Aus Online Mathematik Brückenkurs 2

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}