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Q.

∫012xsin−1x1−x2dx=

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a

12+π23

b

12−π23

c

12+π43

d

12−π43

answer is D.

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Detailed Solution

∫012xsin−11−x2dx,x=sinθ =∫0π6θsinθdθ=−θcosθ+sinθ|0π6 =−π632+12=12−π43
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