Question

The value of $\underset{0}{\overset{2\pi }{\int }}\left[\sin 2x\right(1+\cos 3x\left)\right]dx$, where $\left[t\right]$ denotes the greatest integer function, is :

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

Answer 1

The correct option is D $-\pi$

$I=\underset{0}{\overset{2\pi }{\int }}\left[\sin 2x\right(1+\cos 3x\left)\right]dx...\left(1\right)$
Applying $\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$
$\Rightarrow I=\underset{0}{\overset{2\pi }{\int }}[-\sin 2x(1+\cos 3x\left)\right]dx...\left(2\right)$

We know that, $\left[x\right]+[-x]=-1,x\notin Z$

Adding eqn $\left(1\right)$ and $\left(2\right)$
$\Rightarrow 2I=\underset{0}{\overset{2\pi }{\int }}-1dx$
$\Rightarrow 2I=-2\pi \Rightarrow I=-\pi$