IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

sahyasdev

1. Let F(x)=f(x)+f(x) where f(x)= _{[1 ]}

^{[x ]} iog t dt/1+t . Then prove that F(e) is equal to 1/2

2. Show that

_{[ 0]}

^{[ ]} ^

/2 sec^2x dx/(secx + tan x)^n (n>1)= n/n^2-1

iitkgp_bipin

F(x) = _[1]

^[x] {logt / (1 + t)}.dt + _[1]

^[1/x] {log(t) / (1 + t)}.dt

Differentiating it wrt x :

F'(x) = {logx/(1+x)}.dx/dx + {log(1/x) / (1 + 1/x)}.d(1/x)/dx

F'(x) = {logx/(1+x)} + {logx/(1+x).x}

F'(x) = logx/x

F(x) =

(logx/x).dx

Differentiating it by parts we get :

F(x) = (logx)² -

(logx/x).dx + c = (logx)² - F(x) + c

2F(x) = (logx)² + c

From the original equation at x=1, F(x) = 0.

so, 0 = (log1)² + c

c = 0

so F(x) = (logx)²/2

F(e) = (loge)²/2 = 1/2

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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