Question

The value of ${\int }_0^{2\pi }[\sin 2x(1+\cos 3x)]dx$ , where [t] denotes the greatest integer function, is :

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

Answer 1

Answer: (C)

$I={\int }^{2\pi }\left[\sin 2x\left(1+\cos 3x\right)\right]dx$
$I={\int }_0^{\pi }\left(\left[\sin 2x+\sin 2x\cos 3x\right]+\left[-\sin 2x-\sin 2x\cos 3x\right]\right)dx$
$={\int }_0^{\pi }-dx=-\pi$