If In=∫01 cos−1⁡ xndx then I6−360I2 is given by

Integrating by parts, we obtainIn=∫01 cos−1⁡ xndx=xcos−1⁡ x01+∫01 ncos−1⁡ xn−1x1−x2dx=n∫01 x1−x2cos−1⁡ xn−1dx=n−1−x2cos−1⁡ xn-101−∫01 (n−1)cos-1⁡ xn-2dx=nπ2n−1−n(n−1)In−2I6=6⋅π25−6.5I4=6⋅π25−304π23−12I2I6−360I2=6⋅π25−120π23

If In=∫01 cos−1⁡ xndx then I6−360I2 is given by

Integrating by parts, we obtainIn=∫01 cos−1⁡ xndx=xcos−1⁡ x01+∫01 ncos−1⁡ xn−1x1−x2dx=n∫01 x1−x2cos−1⁡ xn−1dx=n−1−x2cos−1⁡ xn-101−∫01 (n−1)cos-1⁡ xn-2dx=nπ2n−1−n(n−1)In−2I6=6⋅π25−6.5I4=6⋅π25−304π23−12I2I6−360I2=6⋅π25−120π23

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If $I_{n} = \int_{0}^{1} {(\cos^{- 1} x)}^{n} d x$ then $I_{6} - 360 I_{2}$ is given by

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$6 {(\frac{π}{2})}^{5} - 24 {(\frac{π}{2})}^{3}$

$6 {(\frac{π}{2})}^{5} - 120 {(\frac{π}{2})}^{3}$

$6 {(\frac{π}{2})}^{5}$

$6 {(\frac{π}{2})}^{5} - 4 {(\frac{π}{2})}^{3}$

answer is B.

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Detailed Solution

Integrating by parts, we obtain

$\begin{array}{l} I_{n} = \int_{0}^{1} {(\cos^{- 1} x)}^{n} d x = {x \cos^{- 1} x|}_{0}^{1} + \int_{0}^{1} n \frac{{(\cos^{- 1} x)}^{n - 1} x}{\sqrt{1 - x^{2}}} d x \\ = n \int_{0}^{1} \frac{x}{\sqrt{1 - x^{2}}} {(\cos^{- 1} x)}^{n - 1} d x \\ = n [- {\sqrt{1 - x^{2}} {(\cos^{- 1} x)}^{n - 1}|}_{0}^{1} - \int_{0}^{1} (n - 1) {(\cos^{- 1} x)}^{n - 2} d x] \\ = n {(\frac{π}{2})}^{n - 1} - n (n - 1) I_{n - 2} \\ I_{6} = 6 \cdot {(\frac{π}{2})}^{5} - 6.5 I_{4} \\ = 6 \cdot {(\frac{π}{2})}^{5} - 30 [4 {(\frac{π}{2})}^{3} - 12 I_{2}] \\ I_{6} - 360 I_{2} = 6 \cdot {(\frac{π}{2})}^{5} - 120 {(\frac{π}{2})}^{3} \end{array}$