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