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CBSE Class 12-science Answered

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 | 03 Jul, 2017, 07:32: PM
answered-by-expert 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 | 04 Jul, 2017, 08:59: AM

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