\int_{-\pi}^{\pi}\frac{sin2x}{sinx}dx=?

step by step solution:

Recall the **Double-Angle Identity**:

For an angle **x** , the double-angle identity for sine is:

sin2x=2sinxcosx

Using this formula we can write

\int_{-\pi}^{\pi} \frac{\textcolor{primary}{2\sin{x}\cdot \cos{x}}}{\sin{x}}~dx

By simplifying the above integral we get

\int_{-\pi}^{\pi} (\textcolor{primary}{2\cos{x)}}~dx \implies0

The required value of the given integral is:

\int_{-\pi}^{\pi} \frac{\sin2x}{\sin{x}}=0

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