TU Wien:Analysis UE (diverse)/Übungen SS19/Beispiel 221

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Man berechne:

\int \frac{dx}{2\sin^2x\cos^2x}

Hilfreiches

\cos^2x+\sin^2x=1

\int \frac{dx}{\cos^2x} = \tan x = \frac{\sin x}{\cos x}

\int \frac{dx}{\sin^2x} = -\cot x = \frac{\cos x}{\sin x}

Lösungsvorschlag

\begin{align}
\int \frac{dx}{2\sin^2x\cos^2x}
&= \frac{1}{2} \int \frac{1}{\sin^2x\cos^2x} \,dx \\
&= \frac{1}{2} \int \frac{\sin^2x+\cos^2x}{\sin^2x\cos^2x} \,dx \\
&= \frac{1}{2} \int \frac{\sin^2x}{\sin^2x\cos^2x} + \frac{\cos^2x}{\sin^2x\cos^2x} \,dx \\
&= \frac{1}{2} \int \frac{1}{\cos^2x} + \frac{1}{\sin^2x} \,dx \\
&= \frac{1}{2} \left( \int \frac{1}{\cos^2x} \,dx + \int \frac{1}{\sin^2x} \,dx \right) \\
&= \frac{1}{2} \left( \frac{\sin x}{\cos x} - \frac{\cos x}{\sin x} \right) + c \\
&= \frac{\sin^2x - \cos^2x}{2\sin x\cos x} + c \\
&= \frac{\sin^2x - (1 - \sin^2x)}{2\sin x\cos x} + c \\
&= \frac{2\sin^2x - 1}{2\sin x\cos x} + c
\end{align}