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CBSE Class 12-science Answered

Integration (sin-1 x)2 dx explain in great detail
Asked by haroonrashidgkp | 05 Oct, 2018, 12:14: PM
answered-by-expert Expert Answer
begin mathsize 18px style Using space integration space by space parts space straight I space equals integral open parentheses sin to the power of negative 1 end exponent straight x close parentheses squared space dx let space straight u space equals space open parentheses sin to the power of negative 1 end exponent straight x close parentheses squared straight v equals 1 straight I space equals open parentheses sin to the power of negative 1 end exponent straight x close parentheses squared integral 1 dx minus integral open curly brackets straight d over dx open parentheses sin to the power of negative 1 end exponent straight x close parentheses squared. integral 1. dx close curly brackets dx equals straight x open parentheses sin to the power of negative 1 end exponent straight x close parentheses squared minus integral fraction numerator 2 sin to the power of negative 1 end exponent straight x over denominator square root of 1 minus straight x squared end root end fraction. xdx equals straight x open parentheses sin to the power of negative 1 end exponent straight x close parentheses squared plus integral sin to the power of negative 1 end exponent straight x open parentheses fraction numerator negative 2 straight x over denominator square root of 1 minus straight x squared end root end fraction close parentheses dx equals straight x open parentheses sin to the power of negative 1 end exponent straight x close parentheses squared plus open square brackets sin to the power of negative 1 end exponent straight x integral fraction numerator negative 2 straight x over denominator square root of 1 minus straight x squared end root end fraction dx minus integral open curly brackets open parentheses straight d over dx sin to the power of negative 1 end exponent straight x close parentheses integral fraction numerator negative 2 straight x over denominator square root of 1 minus straight x squared end root end fraction dx close curly brackets dx close square brackets equals straight x open parentheses sin to the power of negative 1 end exponent straight x close parentheses squared plus open square brackets sin to the power of negative 1 end exponent straight x.2 square root of 1 minus straight x squared end root minus integral fraction numerator 1 over denominator square root of 1 minus straight x squared end root end fraction.2 square root of 1 minus straight x squared end root dx close square brackets equals straight x open parentheses sin to the power of negative 1 end exponent straight x close parentheses squared plus 2 square root of 1 minus straight x squared end root sin to the power of negative 1 end exponent straight x minus integral 2 dx equals straight x open parentheses sin to the power of negative 1 end exponent straight x close parentheses squared plus 2 square root of 1 minus straight x squared end root sin to the power of negative 1 end exponent straight x minus 2 straight x plus straight c end style
Answered by Arun | 08 Oct, 2018, 10:42: AM

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