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#