Aus Online Mathematik Brückenkurs 2

The integral is a standard integral, with \displaystyle 5x as the argument of the cosine function. If we therefore substitute \displaystyle u=5x, we obtain the “correct” argument of the cosine,

\displaystyle \int\limits_0^{\pi} \cos 5x\,dx = \left\{\begin{align} u &= 5x\\[5pt] du &= (5x)'\,dx = 5\,dx \end{align}\right\} = \frac{1}{5}\int\limits_0^{5\pi} \cos u\,du\,\textrm{.}

As can be seen, the variable change replaced \displaystyle dx by \displaystyle \tfrac{1}{5}\,du and the new limits of integration become \displaystyle u=5\cdot 0=0 and \displaystyle u=5\cdot \pi = 5\pi\,.

Now, we have a standard integral which we can easily compute,

\displaystyle \frac{1}{5}\int\limits_0^{5\pi} \cos u\,du = \frac{1}{5}\Bigl[\ \sin u\ \Bigr]_0^{5\pi} = \frac{1}{5}( \sin 5\pi -\sin 0) = \frac{1}{5}(0-0) = 0\,\textrm{.}

Note: If we draw the graph of \displaystyle y=\cos 5x, we see also that the area between the curve and x-axis above the x-axis is the same as the area under the x-axis.