Question

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

• A. $\frac{3}{5} \sin3x + \frac{8}{3} \sin^{5} x -\sin x +c $
• B. $\frac{2}{5} \sin 5x - \frac{5}{3} \sin^{3} x - 2 \sin x +c $
• C. $\frac{2}{3} \sin 5x - \frac{8}{3} \sin^{5} x + 4 \sin x +c $
• D. $\frac{2}{5} \sin 5 x -\frac{8}{3} \sin^{3} x + 4 \sin x +c $

Answer 1

Answer: (D)

Given, $ I_{n} =\int \frac{\sin n x}{\sin x} d x \dots$(i)
$ I_{n-2} =\int \frac{\sin (n-2) x}{\sin x} d x \dots$(ii)
Subtracting Eq. (ii) from Eq. (i), we get
$I_{n}-I_{n-2}=\int \frac{\{\sin n x-\sin (n-2) x\}}{\sin x} d x$
$=\int \frac{2 \cos (n-1) x \sin x}{\sin x} d x=\int 2 \cos (n-1) x d x$
$=\frac{2 \sin (n-1) x}{(n-1)}$
$\therefore I_{6}-I_{4}=\frac{2 \sin 5 x}{5}$ and $I_{4}-I_{2}=\frac{2 \sin 3 x}{3}$
Now, $ 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 $ I_{6}=I_{4}+2 \frac{\sin 3 x}{5}$
$=I_{2}+2 \frac{\sin 3 x}{3}+2 \frac{\sin 5 x}{5}$
$=2 \frac{\sin 5 x}{5}+2 \frac{\sin 3 x}{3}+2 \sin x+c$
$=\frac{2 \sin 5 x}{5}+\frac{2}{3}\left(3 \sin x-4 \sin ^{3} x\right)+2 \sin x+c$
$I_{6}=\frac{2}{5} \sin 5 x-\frac{8}{3} \sin ^{3} x+4 \sin x+c$