Lösung 2.2:3b

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Version vom 10:22, 11. Mär. 2009

If we are to succeed in simplifying the integral with a substitution, we must find an expression \displaystyle u = u(x) so that the integral can be written as

\displaystyle \int \left(\begin{matrix}

\text{something}\\ \text{in u} \end{matrix}\right)\cdot {u}'\,dx\,\textrm{.}

As our integral is written,

\displaystyle \int\sin x\cos x\,dx

we see that the second factor \displaystyle \cos x is a derivative of the first factor, \displaystyle \sin x. If \displaystyle u=\sin x, the integral can thus be written as

\displaystyle \int u\cdot u'\,dx

and this makes \displaystyle u=\sin x an appropriate substitution,

\displaystyle \begin{align}

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