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Question Number 137467 by SOMEDAVONG last updated on 03/Apr/21

Answered by Ñï= last updated on 03/Apr/21

I=∫(e^(−asin^(−1) x) /( (√(x^2 −1))))dx=∫(e^(−asin^(−1) x) /(i(√(1−x^2 ))))dx=(1/i)∫e^(−asin^(−1) x) d(sin^(−1) x) =−(1/(ia))e^(−asin^(−1) x) +C=(i/a)e^(−asin^(−1) x) +C

I=∫e−asin−1xx2−1dx=∫e−asin−1xi1−x2dx=1i∫e−asin−1xd(sin−1x)=−1iae−asin−1x+C=iae−asin−1x+C

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