Mathematics

# The value of $\displaystyle \int^{\pi}_{0} \dfrac{\sin 2n \,x}{\sin x}dx$ is

$0$

##### SOLUTION
$I=\int ^\pi_0\cfrac{\sin 2nx}{\sin x}dx$
$I=\int^b_a f(x) dx=\int^b_a f(a+b-x)dx$
So, $I=\int ^\pi_0\cfrac{\sin 2n(\pi-x)}{\sin (\pi-x)}dx$
$I=\int ^\pi_0\cfrac{\sin 2nx}{-\sin x}dx$
Adding above  two equations we get
$I+I=$$\int ^\pi_0\cfrac{\sin 2nx}{\sin x}dx$$+\int ^\pi_0\cfrac{\sin 2nx}{-\sin x}dx$
$\implies 2I=0$
$\implies I=0$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

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solve:
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1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
$\displaystyle \int \frac{(x^2-1)}{(x^4+3x^2+1) \tan ^{-1}\left ( \dfrac{x^2+1}{x} \right )}dx=\ln \left | f(x) \right |+C$ then  f(x) is
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Q5 Passage Hard

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