JEE MAIN - Mathematics (2019 - 10th April Morning Slot - No. 18)
The value of $$\int\limits_0^{2\pi } {\left[ {\sin 2x\left( {1 + \cos 3x} \right)} \right]} dx$$,
where [t] denotes the greatest integer function is :
where [t] denotes the greatest integer function is :
2$$\pi $$
$$\pi $$
-2$$\pi $$
-$$\pi $$
Explanation
$$I = \int\limits_0^{2\pi } {\left[ {\sin 2x(1 + \cos 3x)} \right]} dx$$ .... (i)
$$ \therefore\int\limits_0^a {f(x)} = \int\limits_0^a {f(a - x)} dx$$
$$ \because I = \int\limits_0^{2\pi } {\left[ { - \sin 2x(1 + \cos 3x)} \right]} dx$$
By (i) + (ii)
$$ \Rightarrow 2I = \int\limits_0^{2\pi } {( - 1)dx} $$
$$ \Rightarrow 2I = - (x)_0^{2\pi }$$
$$ \Rightarrow $$ I = –$$\pi $$
$$ \therefore\int\limits_0^a {f(x)} = \int\limits_0^a {f(a - x)} dx$$
$$ \because I = \int\limits_0^{2\pi } {\left[ { - \sin 2x(1 + \cos 3x)} \right]} dx$$
By (i) + (ii)
$$ \Rightarrow 2I = \int\limits_0^{2\pi } {( - 1)dx} $$
$$ \Rightarrow 2I = - (x)_0^{2\pi }$$
$$ \Rightarrow $$ I = –$$\pi $$
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