IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

ramkumar_november

solution for the second problem.............

let I =

(secx)^2 dx                                  from 0 to pi/2
         ---------------------------------
           (secx + tanx)^n

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

dt = (secxtanx + (tanx)^2)dx

dt = secx(secx+tanx)dx

dt/t = secxdx.............(ii)

(secx)^2 - (tanx)^2 = 1

[secx+tanx][secx-tanx]=1

[secx-tanx]=1/t...............(iii)

adding (i) and (iii) . we get

2secx = t+ 1/t

=> secx = (t² + 1)/2t........(iv)

therefore the integral becomes

I =

(t² + 1)dt                from 1 to infinity
    ------------------
      tⁿ * 2t * t

I =1/2[

t^-n dt +

t^-n-2dt ]     from 1 to infinity

I = 1/2[    [t^1-n / (1-n)] + [ t^-n-1 /(-1-n)] ]   from 1 to infinity

applying the limits we get........

I = 1/2[   -1/(1-n)    +    1/(n+1) ]

I= 1/2 [ 1/(n+1) - 1/ (1-n)   ]

taking LCM we get

I = 1* [   1-n -n -1   ]
     -----------------------------
      2 * (n+1)*(1-n)

I =     2*n
    ------------------------
     2 * (n+1) * (n-1)

thus we get........

I = n/(n² - 1)

i hope you are clear with my proof>>>>>>>>>>
please do rate me .......................

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