Question

Integration (0 to pi/2) x/1+sinx+cosx

Answer 1

dear friends
I give you some steps for solving question .
step1:- you know from definite integration rule
integ f (x)= integ f (a+b-x)
where a and b is upper and lower limit
so use this
i.e I=integ x/(1+sinx+cosx)
I=integ (pi/2-x)/(1+sinx+cosx)
add this
2I=pi/2. integ 1/(1+sinx+cosx)

step2:- we find
4I/pi= integ 1/(1+sinx+cosx)
put sinx=2tanx/2/(1+tan^2x/2)
and cosx= (1-tan^2x/2)/(1+tan^2x/2)

step3:-we find
4I/pi=integ sec^2x/2/(2+2tanx/2)
let tanx/2=z
defferentiate
1/2. sec^2x/2.dx=dz
now put dx=2.dz/sec^2x/2
step4:-
we find
4I/pi= integ dz/(1+z)
ie. 4I/pi = ln (1+z)
I =pi/4 ln(1+z)
now put z=tanx/2
I=pi/4 ln (1+tanx/2)
put limit (0----> pi/2)
I=(pi/4).ln2