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I am facing several doubts in integration. I have taken picture of two of my Recent doubts. Please help me find out the solution

I am facing several doubts in integration. I have taken picture of two of my Recent doubts. Please help me find out the solution

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Grade:12th pass

1 Answers

neelesh kumar
39 Points
6 years ago
b):     ∫(sec^2x/tanx(tan^2x-1)^(1/2)  )dx
     let tanx  =t
differentiating both side
    we get
(sec^2x) dx = dt
     by puting these values in above integral
we get
     =∫dt/(t(t^2 – 1)^(1/2))
we know that
       d( – cosec^(-1)x)/dx  =  1/x((x^2)   -1)^(1/2)........................(1)
or
     d(sec^(-1)x)/dx  =  1/x((x^2)  -1)^(1/2)..................................(2)
by using one of both 
       we get
{ -  cosec^(-1)t  +c.........................................t  =  tanx        ( when we use    ….1....... )  }
so,,
     {  –  cosec^(-1)tanx + c     }
or when we use...........................2......
        sec^(-1)t  + c                                     t = tanx
{  sec^(-1)tanx +c     }
its ur required  solution

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