Question

How do you find definite Integral?

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

Answer 1

`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`