JEE Main 2019 — Math solution

JEE Main2019Math

The value of $0 \int 2 π [sin 2 x (1 + cos 3 x)] d x,$ where [t] denotes the greatest integer function, is:

Solution

I = \mathop \smallint \limits_0^{2\pi } [\sin 2x(1 + \cos 3x)]dx\,\,\,\,\,\,\,...(i)

$∵$ \mathop \smallint \limits_0^a f(x) = \mathop \smallint \limits_0^a f(a - x)dx

\therefore\,\,\,\,\,\, I = \mathop \smallint \limits_0^{2\pi } [ - \sin 2x(1 + \cos 3x)]dx\,\,\,\,\,\,\,\,\,\,...(ii)

$B y (i) + (ii)$

2I = \mathop \smallint \limits_0^{2\pi } ( - 1)dx

$2 I = - (x)_{0}^{2 π}$

$\Rightarrow I = - π$

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