The value of
Question:

The value of $\int_{0}^{2 \pi}[\sin 2 x(1+\cos 3 x)] d x$, where $[t]$

denotes the greatest integer function, is :

  1. $-2 \pi$

  2. $\pi$

  3. $-\pi$

  4. $2 \pi$


Correct Option: , 3

Solution:

$I=\int_{0}^{2 \pi}[\sin 2 x(1+\cos 3 x)] d x$

$I=\int_{0}^{\pi}([\sin 2 x+\sin 2 x \cos 3 x]+[-\sin 2 x-\sin 2 x \cos 3 x]) d x$

$=\int_{0}^{\pi}-\mathrm{dx}=-\pi$

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