Integrate : under root of tan x with respect to x

Asked by nikhil | 21st Nov, 2015, 08:49: AM

Expert Answer:

integral square root of tan x end root d x S U b s t i t u t e space tan x equals t squared rightwards double arrow s e c squared x d x equals 2 t d t rightwards double arrow open parentheses 1 plus t to the power of 4 close parentheses d x equals 2 t d t rightwards double arrow d x equals fraction numerator 2 t d t over denominator 1 plus t to the power of 4 end fraction S o space t h e space i n t e g r a l space b e c o m e s integral fraction numerator t.2 t d t over denominator 1 plus t to the power of 4 end fraction equals integral fraction numerator 2 t squared d t over denominator 1 plus t to the power of 4 end fraction equals integral fraction numerator t squared plus 1 over denominator 1 plus t to the power of 4 end fraction d t plus integral fraction numerator t squared minus 1 over denominator 1 plus t to the power of 4 end fraction d t D i v i d i n g space n o t h space n u m e r a t o r s space a n d space d e n o m i n a t o r s space b y space apostrophe t squared apostrophe comma space w e space g e t integral fraction numerator 1 plus begin display style 1 over t squared end style over denominator t squared plus begin display style 1 over t squared end style end fraction d t plus integral fraction numerator 1 minus begin display style 1 over t squared end style over denominator t squared plus begin display style 1 over t squared end style end fraction d t equals integral fraction numerator 1 plus begin display style 1 over t squared end style over denominator open parentheses t minus begin display style 1 over t end style close parentheses squared plus 2 end fraction d t plus integral fraction numerator 1 minus begin display style 1 over t squared end style over denominator open parentheses t plus begin display style 1 over t end style close parentheses squared minus 2 end fraction d t S u b s t i t u t e space t minus 1 over t equals u space i n space t h e space f i r s t space i n t e g r a l space a n d space t plus 1 over t equals v space i n space t h e space s e c o n d space i n t e g r a l rightwards double arrow open parentheses 1 plus 1 over t squared close parentheses d t equals d u space a n d space open parentheses 1 minus 1 over t squared close parentheses d t equals d v H e n c e comma space t h e space i n t e g r a l space b e c o m e s integral fraction numerator d u over denominator u squared plus 2 end fraction plus integral fraction numerator d v over denominator v squared minus 2 end fraction equals fraction numerator 1 over denominator square root of 2 end fraction tan to the power of negative 1 end exponent fraction numerator u over denominator square root of 2 end fraction plus fraction numerator 1 over denominator 2 square root of 2 end fraction log open vertical bar fraction numerator v minus square root of 2 over denominator v plus square root of 2 end fraction close vertical bar plus C H e r e space u equals space t minus 1 over t space a n d space v equals t plus 1 over t a n d space t equals square root of tan x end root S u b s t i t u t i n g space b a c k space t h e space v a l u e s comma space w e space g e t space t h e space i n t e g r a l space i n space o r i g i n a l space v a r i a b l e space apostrophe x apostrophe.

Answered by satyajit samal | 25th Nov, 2015, 11:55: AM

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