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Question
Evaluate ∫xcos−1x√1−x2dx
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Solution
I=∫xcos−1x√1−x2dx
Let cos−1x=t,
then, 1√1−x2dx=−dt
I=−∫tcostdt
=−tsint+∫sintdt
=−tsint−cost+C
=−cos−1xsin(cos−1x)−x+C
I=∫xcos−1x√1−x2dx
Let cos−1x=t, then, 1√1−x2dx=−dt
I=−∫tcostdt
=−tsint+∫sintdt
=−tsint−cost+C
=−cos−1xsin(cos−1x)−x+C
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