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∫sec2x√tanx .dx


  • ∫sec2x√tanx .dx

Grade:12

1 Answers

Pratham Ashish
17 Points
12 years ago

hi.

I = ∫sec2x√tanx .dx

 let tan x = t^2

 then, sec^2x dx = 2t dt

        (1+ t^4)  dx = 2t dt

                    dx = 2t / (1+ t^4)   dt

  & cos 2x = 2 cos^2  x   -  1

               = 2 ( 1/  1+ t^4 )  ^2    -1  

                = 2 /  1+ t^4   -1

                =  1- t^4 /  1+ t^4

so   , sec 2x  =  1+ t^4 / 1- t^4

  put these values in I

   I = ∫   { 1+ t^4 / 1- t^4 } . t . {2t / (1+ t^4) } dt

       = ∫ 2 t^2 / 1- t^4    dt

 which can easily be evaluated by partial fraction

 

 

 

 

 

 

 

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