\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}}~dxBy simplifying the above integral we get
\int_{-\pi}^{\pi} (\textcolor{primary}{2\cos{x)}}~dx \implies0The required value of the given integral is:
\int_{-\pi}^{\pi} \frac{\sin2x}{\sin{x}}=0You may like:
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