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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

 

Answers (1)

best_answer

 

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

Posted by

divya.saini

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