Mathematics

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


ANSWER

$$0$$


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
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