How do you find definite Integral?

#int_0^(pi/2) ((sqrt(tanx))/(sqrt(tanx)+sqrt(cotx)))#

1 Answer
May 4, 2018

#I=pi/4#

Explanation:

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#