Value of $\int_{π}^{2 π} [2 s i n x] d x$ where [ ] represents the greatest integer function is

\frac{5 π}{3}

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- π

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\frac{π}{3}

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- 2 π

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Solution

The correct option is A $\frac{5 π}{3}$
$\begin{matrix} l e t I = \int_{π}^{2 π} [2 sin x] d x = - [1 (\frac{7 π}{6} - π) + 2 (\frac{11 π}{6} + \frac{7 π}{6}) + i (2 π - \frac{11 π}{6})] = \frac{- 5 π}{6} \end{matrix}$
$\begin{matrix} f o r π ⩽ x ⩽ 2 π - 2 ⩽ 2 sin x ⩽ 0 w h e n 2 sin = - 1 \Rightarrow sin x = - \frac{1}{2} ∴ x = \frac{7 π}{6}, \frac{11 π}{6} i f - 2 ⩽ 2 sin x < - 1 \Rightarrow \frac{7 π}{6} < x < \frac{11 π}{6} \end{matrix}$
$\begin{matrix} N o w, I = \int \frac{A (6 x + 5) + B}{3 x^{2} + 5 x + 2} d x \Rightarrow I = \int \frac{\frac{5}{6} (6 x + 5) - \frac{1}{6}}{3 x^{2} + 5 x + 2} d x \Rightarrow I = \frac{5}{6} \int \frac{6 x + 5}{3 x^{2} + 5 x + 2} d x - \frac{1}{6} \int \frac{1}{3 x^{2} + 5 x + 2} d x \Rightarrow I = \frac{5}{6} log ∣ ∣ 3 x^{2} + 5 x + 2 ∣ ∣ - \frac{1}{6} I_{1}____(1) \end{matrix}$
Where,
$\begin{matrix} I_{1} = \int \frac{1}{3 x^{2} + 5 x + 2} d x = \int \frac{1}{3 (x^{2} + \frac{5}{3} x + \frac{2}{3})} d x = \frac{1}{3} \int \frac{1}{x^{2} + 2 \times x \times \frac{5}{6} + {(\frac{5}{6})}^{2} - {(\frac{5}{6})}^{2} + \frac{2}{3}} d x = \frac{1}{3} \int \frac{1 d x}{{(x + \frac{5}{6})}^{2} + {(\frac{5}{6})}^{2} + \frac{2}{3}} \end{matrix}$

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