Question

Integrate:

${\int }_0^{\pi /16}\frac{(\sin x+\sin 2x....\sin 7x)}{(\cos x+\cos 2x+....\cos 7x)}dx$

Answer 1

${\int }_0^{\pi /16}\frac{(\sin x+\sin 7x)+(\sin 2x+\sin 6x)+(\sin 3x+\sin 5x)+\sin 4}{(\cos x+\cos 7x)+(\cos 2x+\cos 6x)+(\cos 3x+\cos 5x)+\cos 4x}dx$

$=\int \frac{2\sin 4x.\cos 3x+2\sin 4x.\cos 2x+2\sin 4x.\cos x+\sin 4x}{2\cos 4x.\cos 3x+2\cos 4x.\cos 2x+2\cos 4x.\cos x+\cos 4x.}$

$=\int \frac{\sin 4x}{\cos 4x}\frac{[2\cos 3x+2\cos 2x+2\cos x+1]}{[2\cos 3x+2\cos 2x+2\cos x+1]}$

$={\int }_0^{\pi /16}\tan 4xdx=\frac{1}{4}[-|x|\cos 4x|{]}_0^{\pi /16}$

$=\frac{-1}{4}[ln|\cos \frac{\pi }{4}|-ln|\cos 0|]=\frac{-1}{4}\left[ln\right[1/\sqrt{2}]-ln[1\left]\right]$

$=\frac{-1}{4}\left[ln\right(0.707)-ln[1\left]\right]=\frac{-1}{4}[-0.3467]$

$=\mathrm{0.0866.}$