If ∫sin−1⁡xcos−1⁡xdx=f−1(x)Ax−xf−1(x)−21−x2+π21−x2+2x+c then

I=∫sin-1xcos-1xdx=∫sin−1⁡xπ2−sin−1⁡xdx=π2∫sin−1⁡xdx−∫sin−1⁡x2dx=π2xsin−1⁡x+1−x2−x⋅sin−1⁡x2+2sin−1⁡x1−x2−2x+C=sin−1⁡xπ2x−xsin−1⁡x−21−x2+π21−x2+2x+Cf−1(x)=sin−1⁡x⇒f(x)=sin⁡x

If ∫sin−1⁡xcos−1⁡xdx=f−1(x)Ax−xf−1(x)−21−x2+π21−x2+2x+c then

I=∫sin-1xcos-1xdx=∫sin−1⁡xπ2−sin−1⁡xdx=π2∫sin−1⁡xdx−∫sin−1⁡x2dx=π2xsin−1⁡x+1−x2−x⋅sin−1⁡x2+2sin−1⁡x1−x2−2x+C=sin−1⁡xπ2x−xsin−1⁡x−21−x2+π21−x2+2x+Cf−1(x)=sin−1⁡x⇒f(x)=sin⁡x

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$If \int \sin^{- 1} x \cos^{- 1} x d x = f^{- 1} (x) [A x - x f^{- 1} (x) - 2 \sqrt{1 - x^{2}}] + \frac{π}{2} \sqrt{1 - x^{2}} + 2 x + c then$

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f(x)=sinx

$A = \frac{π}{4}$

$f (x) = \cos x$

$A = \frac{π}{2}$

answer is A, D.

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Detailed Solution

$I = \int \sin^{- 1} x \cos^{- 1} x d x$
$\begin{array}{l} = \int \sin^{- 1} x (\frac{π}{2} - \sin^{- 1} x) d x \\ = \frac{π}{2} \int \sin^{- 1} x d x - \int {(\sin^{- 1} x)}^{2} d x \end{array}$
$= \frac{π}{2} [x \sin^{- 1} x + \sqrt{1 - x^{2}}] - [x \cdot {(\sin^{- 1} x)}^{2} + 2 \sin^{- 1} x \sqrt{1 - x^{2}} - 2 x] + C$

$\begin{array}{l} = \sin^{- 1} x (\frac{π}{2} x - x \sin^{- 1} x - 2 \sqrt{1 - x^{2}}) + \frac{π}{2} \sqrt{1 - x^{2}} + 2 x + C \\ f^{- 1} (x) = \sin^{- 1} x \Rightarrow f (x) = \sin x \end{array}$