# 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}