# The value of the integral $\int_{0}^{\pi}\frac{\sin (2kx)}{\sin x}dx \, \, \, where\, \, \, k\epsilon I\, \, is$ Option 1) $\frac{\pi}{2}$ Option 2) $\pi$ Option 3) 0 Option 4) None of these

Properties of Definite Integration -

$\int_{0}^{2a}f(x)dx= \int_{0}^{a}\left [ f(x)+f(-x) \right ]dx$

$= \left\{\begin{matrix} 2\int_{0}^{a}f(x)dx & if f(2a-x)=f(x)\\ 0 &if f(2a-x)=-f(x) \end{matrix}\right.$

-

$\int_{0}^{\Pi }\frac{sin(2kx)}{sinx}dx$                    K=E

$Let I=\int_{0}^{\Pi }\frac{sin(2kx)}{sinx}dx$

$So I=\int_{0}^{\Pi }\frac{sin(2k(\Pi -x)dx}{sin(\Pi -x)}$                                        $[Put x\rightarrow \Pi -x]$

$I=\int_{0}^{\pi }\frac{sin[2k\pi -2kx]}{sinx}dx = \int_{0}^{\pi }\frac{-sin2kx}{sinx}dx = -I$

$2I=0, I=0$

Option 1)

$\frac{\pi}{2}$

Incorrect

Option 2)

$\pi$

Incorrect

Option 3)

0

Correct

Option 4)

None of these

Incorrect

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