$\int ({sin}^{- 1} x + {cos}^{- 1} x)^{2} d x =$

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\frac{π^{2} x}{4} + c

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\frac{π x}{4} + c

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\frac{π x}{2} + c

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Solution

The correct option is C $\frac{π^{2} x}{4} + c$
We know that
${sin}^{- 1} x + {cos}^{- 1} x = \frac{π}{2}$ .
Hence
$\int ({sin}^{- 1} x + {cos}^{- 1} x)^{2} d x$
$= \int {(\frac{π}{2})}^{2} d x$
$= \int \frac{π^{2}}{4} d x$
$= \frac{π^{2}}{4} \int d x$
$= \frac{π^{2} x}{4} + c$

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