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If $f (x) = \int_{0}^{\sin^{2} x} \sin^{- 1} \sqrt{t} d t$ and $g (x) =$
$\int_{0}^{\cos^{2} x} \cos^{- 1} \sqrt{t}$ dt then the value of $f (x) + g (x)$

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a

π

b

π/4

c

π/2

d

sin2⁡x+sin⁡x+x

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detailed solution

Correct option is B

f′(x)+g′(x)=sin−1⁡(sin⁡x)2sin⁡xcos⁡x−cos−1⁡(cos⁡x)2sin⁡xcos⁡x=xsin⁡2x−xsin⁡2x=0for all x∈R. Hence f(x)+g(x)= constant =C (say) Putting x=π/4,C=∫01/2 sin−1⁡tdt+∫01/2 cos−1⁡tdt=∫01/2 π2dt=π4Hence f(x)+g(x)=π/4

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