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. (1) $\pi$

  2. (2) $-\pi$

  3. (3) $-2 \pi$

  4. (4) $2 \pi$


Correct Option: , 2

Solution:

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

$\because \int_{0}^{a} f(x)=\int_{0}^{a} f(a-x) d x$

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

From (1) + (2), we get;

$2 I=\int_{0}^{2 \pi}(-1) d x \Rightarrow 2 I=-(x)_{0}^{2 \pi} \Rightarrow I=-\pi$

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