∫1/21 Sin−1⁡xdx=

Put x=t. Then 12x⋅dx=dt⇒dx=2tdt;x=1/2,1⇒t=1/2,1∫1/21 Sin−1⁡xdx=2∫1/21 tSin−1⁡tdt=2t2Sin−1⁡t21/21−2∫1/21 11−t2t22dt =π2−π8+∫1/21 1−t2−11−t2dt=3π8+t21−t2+12Sin−1⁡t−Sin−1⁡t121 =3π8+12⋅π2−π2−122⋅12−12⋅π4+π4=3π8+π4−π2−14−π8+π4=π4−14=π−14

∫1/21 Sin−1⁡xdx=

Put x=t. Then 12x⋅dx=dt⇒dx=2tdt;x=1/2,1⇒t=1/2,1∫1/21 Sin−1⁡xdx=2∫1/21 tSin−1⁡tdt=2t2Sin−1⁡t21/21−2∫1/21 11−t2t22dt =π2−π8+∫1/21 1−t2−11−t2dt=3π8+t21−t2+12Sin−1⁡t−Sin−1⁡t121 =3π8+12⋅π2−π2−122⋅12−12⋅π4+π4=3π8+π4−π2−14−π8+π4=π4−14=π−14

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$\int_{1 / 2}^{1} {Sin}^{- 1} \sqrt{x} dx =$

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$\frac{π - 1}{4}$

$\frac{π - 1}{8}$

$\frac{π - 2}{4}$

$\frac{π - 2}{8}$

answer is A.

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Detailed Solution

Put $\sqrt{x} = t . Then \frac{1}{2 \sqrt{x}} \cdot dx = dt \Rightarrow dx = 2 tdt; x = 1 / 2, 1 \Rightarrow t = 1 / \sqrt{2}, 1$
$\begin{array}{l} \int_{1 / 2}^{1} {Sin}^{- 1} \sqrt{x} dx = 2 \int_{1 / \sqrt{2}}^{1} {tSin}^{- 1} tdt \\ = 2 {[\frac{t^{2} {Sin}^{- 1} t}{2}]}_{1 / \sqrt{2}}^{1} - 2 \int_{1 / \sqrt{2}}^{1} \frac{1}{\sqrt{1 - t^{2}}} \frac{t^{2}}{2} dt \\ = \frac{π}{2} - \frac{π}{8} + \int_{1 / \sqrt{2}}^{1} [\sqrt{1 - t^{2}} - \frac{1}{\sqrt{1 - t^{2}}}] dt \\ = \frac{3 π}{8} + {[\frac{t}{2} \sqrt{1 - t^{2}} + \frac{1}{2} {Sin}^{- 1} t - {Sin}^{- 1} t]}_{1 \sqrt{2}}^{1} \\ = \frac{3 π}{8} + \frac{1}{2} \cdot \frac{π}{2} - \frac{π}{2} - \frac{1}{2 \sqrt{2}} \cdot \frac{1}{\sqrt{2}} - \frac{1}{2} \cdot \frac{π}{4} + \frac{π}{4} \\ = \frac{3 π}{8} + \frac{π}{4} - \frac{π}{2} - \frac{1}{4} - \frac{π}{8} + \frac{π}{4} = \frac{π}{4} - \frac{1}{4} = \frac{π - 1}{4} \end{array}$