IIT JEE , AIEEE Forum - Mathematics , Integral Calculus - Integrals Challenge : rates for best answer

ramkumar_november

guys , i found out another method....... check whether its correct........

I =

dx
-----------------------------------
1 + sinx + sin²x

I =

      sec²x   dx
         -----------------------------------------------
           sec²x   + tan²x   + secxtanx

I =

   secx.secx.dx
       -----------------------------------------------------
         (secx + tanx)²   - secxtanx

now put secx + tanx = t ..................(i)

           secx(secx+tanx)dx=dt

              secxdx=dt/t

also    secx - tanx =1/t . ......................(ii)

adding (i) and(ii)

secx = (t² + 1)/2t    and tanx = (t² - 1)/2t

so secxtanx = (t⁴ - 1)/4t²

I =

   (t²+1)dt
      -----------------------------------------
        t*2t*{ t²   - (t⁴ - 1)/4t² }

I =

2(t²+1)dt
-----------------------
3t⁴ + 1

after which i think you can solve.......................

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