In $I_n=\int \frac{\sin n x}{\sin x} d x$ for $n=1,2,3, \ldots$, then $I_6=$

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Aicharya · Accepted Answer

Subtracting Eq. (ii) from Eq. (i), we get $ \begin{aligned} & I_n-I_{n-2}=\int \frac{\{\sin n x-\sin (n-2) x\}}{\sin x} d x \\ & =\int \frac{2 \cos (n-1) x \sin x}{\sin x} d x=\int 2 \cos (n-1) x d x \\ & =\frac{2 \sin (n-1) x}{(n-1)} \end{aligned} $ $\therefore \quad I_6-I_4=\frac{2 \sin 5 x}{5}$ and $I_4-I_2=\frac{2 \sin 3 x}{3}$ Now, $\quad I_2=\int \frac{\sin 2 x}{\sin x} d x$ $ =\int \frac{2 \sin x \cos x}{\sin x} d x=2 \int \cos x d x $ $ =2 \sin x+c $ and $ \begin{aligned} I_6 & =I_4+2 \frac{\sin 5 x}{5} \\ & =I_2+2 \frac{\sin 3 x}{3}+2 \frac{\sin 5 x}{5} \end{aligned} $ $ \begin{aligned} & =2 \frac{\sin 5 x}{5}+2 \frac{\sin 3 x}{3}+2 \sin x+c \\ & =\frac{2 \sin 5 x}{5}+\frac{2}{3}\left(3 \sin x-4 \sin ^3 x\right)+2 \sin x+c \\ & I_6=\frac{2}{5} \sin 5 x-\frac{8}{3} \sin ^3 x+4 \sin x+c \end{aligned} $

In $I_n=\int \frac{\sin n x}{\sin x} d x$ for $n=1,2,3, \ldots$, then $I_6=$

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