∫Sin−1x−Cos−1xSin−1x+Cos−1xdx=

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4π[xSin–1x+1–x2]−x+c

1π[xSin–1x+1–x2]−x+c

2π[xSin–1x+1–x2]−x+c

2π[xSin–1x–1–x2]−x+c

answer is A.

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Detailed Solution

∫2sin−1x−π2π2dx =2π2∫sin−1xdx−∫1 dx =4π[sin−1x.x−∫x1−x2dx]−x =4πxsin-1x+12∫-2x1-x2dx -x =4πxsin-1x+1221-x2 -x+c using ∫dtt=2t =4π[xsin−1x+1−x2]−x+c

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