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Trigonometric integral

Evaluate the integral

    \[\mathcal{J} =\int \frac{\mathrm{d}x}{\sin x \sin (x+1)}\]

Solution

We have successively:

    \begin{align*} \mathcal{J} &= \int \frac{\mathrm{d}x}{\sin x \sin (x+1)} \\ &= \frac{1}{\sin 1} \int \frac{\sin 1}{\sin x \sin (x+1)} \, \mathrm{d}x \\ &= \frac{1}{\sin 1} \int \frac{\sin \left ( x+1-x \right )}{\sin x \sin (x+1)}\, \mathrm{d}x \\ &= \frac{1}{\sin 1} \int \frac{\sin (x+1) \cos x - \cos (x+1) \sin x}{\sin x \sin (x+1)} \, \mathrm{d}x\\ &= \frac{1}{\sin 1} \int \left ( \frac{\sin (x+1) \cos x}{\sin x \sin (x+1)} - \frac{\cos (x+1) \sin x}{\sin x \sin (x+1)} \right ) \, \mathrm{d}x \\ &= \frac{1}{\sin 1} \int \left ( \frac{\cos x}{\sin x} - \frac{\cos (x+1)}{\sin (x+1)} \right ) \, \mathrm{d}x \\ &= \frac{1}{\sin 1} \left ( \ln \sin x - \ln \sin (x+1) \right ) + c \; , \; c \in \mathbb{R} \end{align*}

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