If I_n = int frac{sin nx}{sin x} dx for n = 1 | vedclass

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(D) Given $I_n = \int \frac{\sin nx}{\sin x} dx$. Consider $I_n - I_{n-2} = \int \frac{\sin nx - \sin(n-2)x}{\sin x} dx$. Using the formula $\sin A -

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$\frac{2}{5} \sin 5x - \frac{8}{3} \sin^3 x + 4 \sin x + c$

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(D) Given $I_n = \int \frac{\sin nx}{\sin x} dx$. Consider $I_n - I_{n-2} = \int \frac{\sin nx - \sin(n-2)x}{\sin x} dx$. Using the formula $\sin A - \sin B = 2 \cos(\frac{A+B}{2}) \sin(\frac{A-B}{2})$,we get: $I_n - I_{n-2} = \int \frac{2 \cos((n-1)x) \sin x}{\sin x} dx = \int 2 \cos((n-1)x) dx = \frac{2 \sin((n-1)x)}{n-1} + C'$. For $n=6$,$I_6 - I_4 = \frac{2 \sin 5x}{5}$. For $n=4$,$I_4 - I_2 = \frac{2 \sin 3x}{3}$. For $n=2$,$I_2 = \int \frac{\sin 2x}{\sin x} dx = \int \frac{2 \sin x \cos x}{\sin x} dx = 2 \int \cos x dx = 2 \sin x + C''$. Summing these,$I_6 = (I_6 - I_4) + (I_4 - I_2) + I_2 = \frac{2 \sin 5x}{5} + \frac{2 \sin 3x}{3} + 2 \sin x + c$. Using $\sin 3x = 3 \sin x - 4 \sin^3 x$: $I_6 = \frac{2}{5} \sin 5x + \frac{2}{3}(3 \sin x - 4 \sin^3 x) + 2 \sin x + c$ $I_6 = \frac{2}{5} \sin 5x + 2 \sin x - \frac{8}{3} \sin^3 x + 2 \sin x + c$ $I_6 = \frac{2}{5} \sin 5x - \frac{8}{3} \sin^3 x + 4 \sin x + c$.

If $I_n = \int \frac{\sin nx}{\sin x} dx$ for $n = 1, 2, 3, \ldots$,then $I_6 =$

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