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CBSE Class 12-science Answered

Kindly provide the complete solution 
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Asked by Varsneya Srinivas | 12 Feb, 2018, 04:58: PM
answered-by-expert Expert Answer
begin mathsize 16px style integral square root of fraction numerator 1 plus straight x over denominator straight x end fraction end root dx

put space
straight x equals tan squared straight y
dx equals 2 space tan space straight y space sec squared straight y space dy
integral square root of fraction numerator 1 plus tan squared straight y over denominator tan squared straight y end fraction end root 2 space tan space straight y space sec squared straight y space dy
integral square root of fraction numerator sec squared straight y over denominator tan squared straight y end fraction end root 2 space tan space straight y space sec squared straight y space dy
integral fraction numerator sec space straight y over denominator tan space straight y end fraction 2 space tan space straight y space sec squared straight y space dy
2 integral sec cubed straight y space dy
2 integral secy space sec squared straight y space dy
apply space by space parts space
secy space equals straight u
sec squared straight y equals straight v
straight I equals 2 open parentheses secy integral sec squared ydy minus integral open square brackets straight d over dx secy integral sec squared ydy close square brackets close parentheses
straight I over 2 equals secy space tany space minus integral secy space space tan space straight y space tan space straight y space dy
equals secy space tany space minus integral secy space space tan squared space straight y space space dy

equals secy space tany space minus integral secy space space open parentheses sec squared straight y minus 1 close parentheses space space dy
straight I over 2 equals secy space tany space minus integral space sec cubed straight y space dy minus integral sec space straight y space space dy
straight I over 2 equals secy space tany space minus straight I minus log open vertical bar secy space plus tany close vertical bar plus straight c
fraction numerator 3 straight I over denominator 2 end fraction equals secy space tany space minus log open vertical bar secy space plus tany close vertical bar plus straight c
straight I equals fraction numerator 2 open parentheses secy space tany space minus log open vertical bar secy space plus tany close vertical bar close parentheses over denominator 3 end fraction plus straight c
re space sub space
straight x equals space tany
secy space equals square root of 1 plus straight x squared end root

end style
Answered by Arun | 12 Feb, 2018, 05:49: PM
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