IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

kuldud

1...integral 0 to pi/4[log(1+tanx)]dx

2...integral 0to1 [sqrt(tanx)]dx

kuldud

common expert guys plzz help me out

if u cn den rates for sure

cheers

try it out

dese r standard sums

do dem n b gain confidence

kuldud

so guys u cant solv dese questions

common yaaer try it out

i need it urgent

rates assurd

chimanshu_007

ans-1)

let I = _{[0 ]}^{[pi/4 ]} log (1+tanx)

then I =

_{[0 ]}

^{[ pi/4]} log [ 1 + tan (pi/4 - x) ] dx

I =₀

^pi/4 log ( 1 + (tan pi/4 - tanx / 1 + tan(pi/4) tanx) )dx

I = ₀

^pi/4log[1 + (1-tanx / 1 + tanx)] dx

= ₀

^pi/4 log [ (1 + tanx + 1 - tanx) / 1 + tanx] dx

= ₀

^pi/4 log (2 / tanx) dx

seperating log

= ₀

^pi/4 { log2 - log (1 + tanx)} dx

= ₀

^pi/4 log2dx - ₀

^pi/4 log(1+tanx) dx

as ₀

^pi/4 log(1+tanx) = I

I = log 2 [ x ]^pi/4_o - I

2I = log 2 ( pi/4)

I = log2 ( pi/8)

this is the answer

venkat_tatolu

For secon problem, put tanx=tsquare.