Find begin mathsize 20px style integral tan to the power of negative 1 end exponent square root of fraction numerator 1 minus x over denominator 1 plus x end fraction end root end style

Asked by sunil2791 | 3rd Jul, 2017, 07:32: PM

Expert Answer:

begin mathsize 16px style straight I equals integral tan to the power of negative 1 end exponent open parentheses square root of fraction numerator 1 minus straight x over denominator 1 plus straight x end fraction end root close parentheses dx
In space these space case space of space integration space of space inverse space function space we space always space try space to space
eliminate space this space inverse space using space substitution space method.
straight x equals cosθ space rightwards double arrow dx equals negative sinθdθ space
straight I equals integral tan to the power of negative 1 end exponent open parentheses square root of fraction numerator 1 minus cosθ over denominator 1 plus cosθ end fraction end root close parentheses open parentheses negative sinθdθ close parentheses
straight I equals negative integral tan to the power of negative 1 end exponent open parentheses square root of fraction numerator 2 sin squared begin display style straight theta over 2 end style over denominator 2 cos squared straight theta over 2 end fraction end root close parentheses sinθdθ
straight I equals negative integral tan to the power of negative 1 end exponent open parentheses tan straight theta over 2 close parentheses sinθdθ
straight I equals negative integral straight theta over 2 sinθdθ
straight I equals fraction numerator negative 1 over denominator 2 end fraction integral θsinθdθ
Use space integration space by space parts space and space solve space it.
end style

Answered by Sneha shidid | 4th Jul, 2017, 08:59: AM