We know that,
If f is c ontinuous in [0,a] then
color(blue)((1)int_0^a f(x)dx=int_0^a f(a-x)dx
Here,
I=int_0^(pi/2) sqrttanx/(sqrttanx+sqrtcotx)dx...to(A)
Using (1) , we get
I= int_0^(pi/2) sqrttan(pi/2-x)/(sqrttan(pi/2-x)+sqrtcot(pi/2-x))dx
I=int_0^(pi/2) sqrtcotx/(sqrtcotx+sqrttanx)dx...to(B)
Adding (A) and (B)
I+I=int_0^(pi/2) sqrttanx/(sqrttanx+sqrtcotx)dx+int_0^(pi/2)
sqrtcotx/(sqrttanx+sqrtcotx)dx
2I=int_0^(pi/2) (sqrttanx+sqrtcotx)/(sqrttanx+sqrtcotx)dx
2I=int_0^(pi/2) 1dx
2I=[x]_0^(pi/2)
2I=pi/2-0=pi/2
=>I=pi/4