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CBSE Class 12-science Answered

Find begin mathsize 20px style integral open parentheses square root of tan x end root plus square root of c o t x end root close parentheses end style
Asked by sunil2791 | 03 Jul, 2017, 07:37: PM
answered-by-expert Expert Answer
begin mathsize 16px style straight I equals integral open parentheses square root of tanx plus square root of cotx close parentheses dx straight I equals integral open parentheses square root of cotx plus fraction numerator 1 over denominator square root of cotx end fraction close parentheses dx straight I equals integral fraction numerator cotx plus 1 over denominator square root of cotx end fraction dx straight I equals integral square root of tanx open parentheses cotx plus 1 close parentheses dx straight I equals integral square root of tanx open parentheses 1 over tanx plus 1 close parentheses dx Put space tanx equals straight t squared space rightwards double arrow sec squared xdx equals 2 tdt rightwards double arrow open parentheses 1 plus tan squared straight x close parentheses dx equals 2 tdt rightwards double arrow dx equals fraction numerator 2 tdt over denominator 1 plus straight t to the power of 4 end fraction straight I equals integral straight t open parentheses 1 over straight t squared plus 1 close parentheses fraction numerator 2 tdt over denominator 1 plus straight t to the power of 4 end fraction straight I equals integral 2 straight t squared open parentheses fraction numerator 1 plus straight t squared over denominator straight t squared end fraction close parentheses fraction numerator dt over denominator 1 plus straight t to the power of 4 end fraction straight I equals 2 integral fraction numerator 1 plus straight t squared over denominator 1 plus straight t to the power of 4 end fraction dt straight I equals integral fraction numerator begin display style fraction numerator 1 plus straight t squared over denominator straight t squared end fraction end style over denominator begin display style fraction numerator 1 plus straight t to the power of 4 over denominator straight t squared end fraction end style end fraction dt straight I equals integral fraction numerator begin display style 1 over straight t squared end style plus 1 over denominator 1 over straight t squared plus straight t squared end fraction dt straight I equals integral fraction numerator begin display style 1 over straight t squared plus 1 end style over denominator open parentheses straight t minus begin display style 1 over straight t end style close parentheses squared plus 2 end fraction dt Put space straight t minus 1 over straight t equals straight z comma space open parentheses 1 over straight t squared plus 1 close parentheses dt equals dz straight I equals integral fraction numerator dz over denominator straight z squared plus open parentheses square root of 2 close parentheses squared end fraction straight I equals tan to the power of negative 1 end exponent open parentheses fraction numerator space straight t minus 1 over straight t over denominator square root of 2 end fraction close parentheses plus straight c straight I equals tan to the power of negative 1 end exponent open parentheses fraction numerator straight t squared minus 1 over denominator square root of 2 straight t end fraction close parentheses plus straight c straight I equals tan to the power of negative 1 end exponent open parentheses fraction numerator tan straight x minus 1 over denominator square root of 2 tanx end root end fraction close parentheses plus straight c end style
Answered by Sneha shidid | 04 Jul, 2017, 09:51: AM

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