Question

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

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

Answer 1

Answer: (D)

Given, $I_n=\int \frac{\sin nx}{\sin x}dx\dots$(i)
$I_{n-2}=\int \frac{\sin (n-2)x}{\sin x}dx\dots$(ii)
Subtracting Eq. (ii) from Eq. (i), we get
$I_n-I_{n-2}=\int \frac{\{\sin nx-\sin (n-2)x\}}{\sin x}dx$
$=\int \frac{2\cos (n-1)x\sin x}{\sin x}dx=\int 2\cos (n-1)xdx$
$=\frac{2\sin (n-1)x}{(n-1)}$
$\therefore I_6-I_4=\frac{2\sin 5x}{5}$ and $I_4-I_2=\frac{2\sin 3x}{3}$
Now, $I_2=\int \frac{\sin 2x}{\sin x}dx$
$=\int \frac{2\sin x\cos x}{\sin x}dx=2\int \cos xdx$
$=2\sin x+c$
and $I_6=I_4+2\frac{\sin 3x}{5}$
$=I_2+2\frac{\sin 3x}{3}+2\frac{\sin 5x}{5}$
$=2\frac{\sin 5x}{5}+2\frac{\sin 3x}{3}+2\sin x+c$
$=\frac{2\sin 5x}{5}+\frac{2}{3}\left(3\sin x-4{\sin }^{3}x\right)+2\sin x+c$
$I_6=\frac{2}{5}\sin 5x-\frac{8}{3}{\sin }^{3}x+4\sin x+c$