Question

The value of integral $\underset{0}{\overset{\pi }{\int }}(1-|\sin 8x|)dx$ is

• A. $0$
• B. $\pi -1$
• C. $\pi -2$
• D. $\pi -3$

Answer 1

The correct option is C $\pi -2$

$LetI=\underset{0}{\overset{\pi }{\int }}(1-|\sin 8x|)dx=\pi -\underset{0}{\overset{\pi }{\int }}|\sin 8x|dx=\pi -8\underset{0}{\overset{\frac{\pi }{8}}{\int }}|\sin 8x|dx=\pi -8\underset{0}{\overset{\frac{\pi }{8}}{\int }}\sin 8xdx=\pi +8{\left[\frac{\cos 8x}{8}\right]}_0^{\frac{\pi }{8}}$

$\Rightarrow I=\pi -2$