Question

${\int }_0^{\pi }1-|\sin 8x|$

• A. $1$
• B. $\pi -1$
• C. $\pi -2$
• D. $\pi -3$

Answer 1

Answer: (C)

$\pi -2$

${\int }_0^{\pi }(1-\left|\sin 8x\right|)dx$

$={\int }_0^{\pi }1.dx{\int }_0^{\pi /8}\sin 8xdx+{\int }_{\pi /8}^{\pi /4}\sin 8xdx-{\int }_{\pi /4}^{3\pi /8}\sin 8xdx+{\int }_{3\pi /8}^{\pi /2}\sin 8xdx-{\int }_{\pi /2}^{5\pi /8}\sin 8xdx+{\int }_{5\pi /8}^{6\pi /8}\sin 8xdx-{}^{}{\int }_{6\pi /8}^{7\pi /8}\sin 8xdx+{\int }_{7\pi /8}^{8\pi /8}\sin 8xdx$

$={\left[x\right]}_0^{\pi }-{\left[\frac{\cos 8x}{8}\right]}_0^{\pi /8}+{\left[\frac{-\cos 8x}{8}\right]}_{\pi /8}^{\pi /4}-{\left[\frac{-\cos 8x}{8}\right]}_{\pi /4}^{3\pi /8}+{\left[\frac{-\cos 8x}{8}\right]}_{3\pi /8}^{\pi /2}-{\left[\frac{-\cos 8x}{8}\right]}_{\pi /2}^{5\pi /8}+{\left[\frac{-\cos 8x}{8}\right]}_{5\pi /8}^{3\pi /4}-{\left[\frac{-\cos 8x}{8}\right]}_{3\pi /4}^{\pi /8}+{\left[\frac{-\cos 8x}{8}\right]}_{7\pi /8}^{\pi }$

$=\pi -\frac{2}{8}-\frac{2}{8}-\frac{2}{8}-\frac{2}{8}-\frac{2}{8}-\frac{2}{8}-\frac{2}{8}-\frac{2}{8}$

$=\pi -2.$

Hence, the answer is $\pi -2.$