The value of ∫−11 sin−1⁡x2+12dx+∫−11 cos−1⁡x2−12dx , where[·] denotes the greatest integer function, is

Let I1=∫−11 sin−1⁡x2+12dxI1=2∫01 sin−1⁡x2+12dxI1=2∫01/2 sin−1⁡x2+12dx+∫1/21 sin−1⁡x2+12dxI1=2∫01/2 sin−1⁡0dx+∫1/21 sin−1⁡1dxI1=2π2∫1/21 dx=π1−12and , I2=∫−11 cos−1⁡x2−12dx ⇒ I2=2∫01 cos−1⁡x2−12dx⇒ I2=2∫01/2 cos−1⁡x2−12dx+∫1/21 cos−1⁡x2−12dx⇒ I2=2∫01/2 cos−1⁡(−1)dx+∫1/21 cos−1⁡0dx⇒ I2=2∫01/2 πdx+∫1/21 π2dx⇒ l2=2π12−0+π21−12⇒ I2=2π122+12=π12+1⇒I1+I2=π1−12+π12+1=2π

The value of ∫−11 sin−1⁡x2+12dx+∫−11 cos−1⁡x2−12dx , where[·] denotes the greatest integer function, is

Let I1=∫−11 sin−1⁡x2+12dxI1=2∫01 sin−1⁡x2+12dxI1=2∫01/2 sin−1⁡x2+12dx+∫1/21 sin−1⁡x2+12dxI1=2∫01/2 sin−1⁡0dx+∫1/21 sin−1⁡1dxI1=2π2∫1/21 dx=π1−12and , I2=∫−11 cos−1⁡x2−12dx ⇒ I2=2∫01 cos−1⁡x2−12dx⇒ I2=2∫01/2 cos−1⁡x2−12dx+∫1/21 cos−1⁡x2−12dx⇒ I2=2∫01/2 cos−1⁡(−1)dx+∫1/21 cos−1⁡0dx⇒ I2=2∫01/2 πdx+∫1/21 π2dx⇒ l2=2π12−0+π21−12⇒ I2=2π122+12=π12+1⇒I1+I2=π1−12+π12+1=2π

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The value of $\int_{- 1}^{1} \sin^{- 1} [x^{2} + \frac{1}{2}] d x + \int_{- 1}^{1} \cos^{- 1} [x^{2} - \frac{1}{2}] d x$ , where[·] denotes the greatest integer function, is

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$π$

$2 π$

$4 π$

answer is B.

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Detailed Solution

Let

$\begin{aligned} I_{1} = \int_{- 1}^{1} \sin^{- 1} [x^{2} + \frac{1}{2}] d x \\ I_{1} = 2 \int_{0}^{1} \sin^{- 1} [x^{2} + \frac{1}{2}] d x \end{aligned}$

$\begin{array}{l} I_{1} = 2 [\int_{0}^{1 / \sqrt{2}} \sin^{- 1} [x^{2} + \frac{1}{2}] d x + \int_{1 / \sqrt{2}}^{1} \sin^{- 1} [x^{2} + \frac{1}{2}] d x] \\ I_{1} = 2 [\int_{0}^{1 / \sqrt{2}} \sin^{- 1} 0 d x + \int_{1 / \sqrt{2}}^{1} \sin^{- 1} 1 d x] \\ I_{1} = 2 [\frac{π}{2} \int_{1 / \sqrt{2}}^{1} d x] = π (1 - \frac{1}{\sqrt{2}}) \end{array}$

and , $I_{2} = \int_{- 1}^{1} \cos^{- 1} [x^{2} - \frac{1}{2}] d x$

$\begin{array}{l} \Rightarrow I_{2} = 2 \int_{0}^{1} \cos^{- 1} [x^{2} - \frac{1}{2}] d x \\ \Rightarrow I_{2} = 2 [\int_{0}^{1 / \sqrt{2}} \cos^{- 1} [x^{2} - \frac{1}{2}] d x + \int_{1 / \sqrt{2}}^{1} \cos^{- 1} [x^{2} - \frac{1}{2}] d x] \\ \Rightarrow I_{2} = 2 [\int_{0}^{1 / \sqrt{2}} \cos^{- 1} (- 1) d x + \int_{1 / \sqrt{2}}^{1} \cos^{- 1} 0 d x] \\ \Rightarrow I_{2} = 2 [\int_{0}^{1 / \sqrt{2}} π d x + \int_{1 / \sqrt{2}}^{1} \frac{π}{2} d x] \\ \Rightarrow l_{2} = 2 [π [\frac{1}{\sqrt{2}} - 0) + \frac{π}{2} (1 - \frac{1}{\sqrt{2}})] \\ \Rightarrow I_{2} = 2 [π (\frac{1}{2 \sqrt{2}} + \frac{1}{2})] = π (\frac{1}{\sqrt{2}} + 1) \\ \Rightarrow I_{1} + I_{2} = π (1 - \frac{1}{\sqrt{2}}) + π (\frac{1}{\sqrt{2}} + 1) = 2 π \end{array}$