i n t e g r a t e space integral sin to the power of negative 1 end exponent square root of fraction numerator x over denominator a plus x end fraction space end root space d x

Asked by manish.kkd4 | 20th Feb, 2019, 12:15: PM

Expert Answer:

begin mathsize 16px style straight I equals integral sin to the power of negative 1 end exponent square root of fraction numerator straight x over denominator straight a plus straight x end fraction end root dx
straight x equals atan squared straight theta
dx equals 2 atanθsec squared θdθ
straight I equals integral sin to the power of negative 1 end exponent open parentheses square root of fraction numerator atan squared straight theta over denominator straight a plus atan squared straight theta end fraction end root close parentheses tanθsec squared θdθ
straight I equals 2 straight a integral sin to the power of negative 1 end exponent open parentheses square root of fraction numerator atan squared straight theta over denominator straight a open parentheses 1 plus tan squared straight theta close parentheses end fraction end root close parentheses tanθsec squared θdθ
straight I equals 2 straight a integral sin to the power of negative 1 end exponent open parentheses square root of fraction numerator tan squared straight theta over denominator sec squared straight theta end fraction end root close parentheses tanθsec squared θdθ
straight I equals 2 straight a integral sin to the power of negative 1 end exponent open parentheses sinθ close parentheses tanθsec squared θdθ
straight I equals 2 straight a integral θtanθ open parentheses 1 plus tan squared straight theta close parentheses dθ
Integrate space further space using space integration space by space parts. end style

Answered by Sneha shidid | 20th Feb, 2019, 01:07: PM