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The value of ∫(sin^-1 x)^2 dx for x ∈ [0,1] is ... (a) (π^2/4) - 1 (b) (π^2/4) + 2 (c) (π^2/4) + 1

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The value of ∫(sin-1 x)2 dx for x ∈ [0,1] is ...

(a) (π2/4) - 1

(b) (π2/4) + 2

(c) (π2/4) + 1

(d) (π2/4) - 2

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Answer is (d) (π2/4) - 2

Put x = sin t

dx = cos t dt 

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