Correct Answer - Option 2 : -π
\({\rm{I}} = \mathop \smallint \nolimits_0^{2{\rm{\pi }}} \left[ {{\rm{sin}}2{\rm{x}}\left( {1 + {\rm{cos}}3{\rm{x}}} \right)} \right]{\rm{dx}}\) ----(1)
\(\mathop \smallint \nolimits_0^{\rm{a}} f\left( {\rm{x}} \right) = \mathop \smallint \nolimits_0^{\rm{a}} f\left( {{\rm{a}} - {\rm{x}}} \right){\rm{dx}}\)
(1) \(\Rightarrow \mathop{\int }_{0}^{2\text{ }\!\!\pi\!\!\text{ }}\left[ \sin \left( 2\text{ }\!\!\pi\!\!\text{ }-2\text{x} \right)\left( 1+\cos \left( 2\text{ }\!\!\pi\!\!\text{ }-3\text{x} \right) \right) \right]\text{dx}\)
\(\text{I}=\mathop{\int }_{0}^{2\text{ }\!\!\pi\!\!\text{ }}\left[ -\text{sin }\!\!~\!\!\text{ }2\text{x}\left( 1+\text{cos }\!\!~\!\!\text{ }3\text{x} \right) \right]\text{dx}\) ----(2)
∵ sin (2π – x) = -sin x
∵ cos (2π – x) = cos x
By adding equation (1) and (2),
\(\Rightarrow {\rm{I}} + {\rm{I}} = \mathop \smallint \nolimits_0^{2{\rm{\pi }}} {\rm{}}\left[ {\left( {{\rm{sin\;}}2{\rm{x}}\left( {1 + {\rm{cos\;}}3{\rm{x}}} \right)} \right) + \left( { - {\rm{sin\;}}2{\rm{x}}\left( {1 + {\rm{cos\;}}3{\rm{x}}} \right)} \right)} \right]{\rm{dx}}\)
By using one of the greatest integer properties, which is:
[x] + [-x] = -1
Now, applying the above property,
\(\Rightarrow 2{\rm{I}} = \mathop \smallint \nolimits_0^{2{\rm{\pi }}} \left( { - 1} \right){\rm{dx}}\)
\(\Rightarrow 2{\rm{I}} = - \left[ {\rm{x}} \right]_0^{2{\rm{\pi }}}\)
⇒ 2I = -2π – 0
\(\Rightarrow {\rm{I}} = \frac{{ - 2{\rm{\pi }}}}{2}\)
∴ I = -π