Aus Online Mathematik Brückenkurs 2

Let's rewrite the integral somewhat:

\displaystyle 2\sin \sqrt{x}\centerdot \frac{1}{2\sqrt{x}}

Here, we see that the factor on the right, \displaystyle \frac{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\centerdot {u}'

and the integral becomes

\displaystyle \begin{align} & \int{\frac{\sin \sqrt{x}}{\sqrt{x}}}\,dx=\left\{ \begin{matrix} u=\sqrt{x} \\ du=\left( \sqrt{x} \right)^{\prime }\,dx=\frac{1}{2\sqrt{x}}\,dx \\ \end{matrix}\, \right\} \\ & =2\int{\sin u\,du} \\ & =-2\cos u+C \\ & =-2\cos \sqrt{x}+C \\ \end{align}