`int_(0)^(pi//2)tan^(-1)[(sqrt(1+sin x)+sqrt(1-sin x))/(sqrt(1+sin x)-sqrt(1-sin x))]\ dx`

cot^(-1)[(sqrt(1+sin x)+sqrt(1-sin x))/(sqrt(1+sin x)-sqrt(1-sin x))]=

(cot^(-1){sqrt(1+sin x)+sqrt(1-sin x)})/(sqrt(1+sin x)-sqrt(1-sin x))

(d)/(dx) Tan^(-1)[(sqrt(1+sinx) - sqrt(1-sin x))/(sqrt(1+sin x) + sqrt(1-sin x))]=

cot^(-1)((sqrt(1+sin x)+sqrt(1-sin x))/(sqrt(1+sin x)-sqrt(1-sin x)))=(x)/(2)

Differentiate tan^(-1) ((sqrt(1 + sin x) + sqrt(1 - sin x))/(sqrt(1 + sin x) - sqrt(1 - sin x))) w.r.t.x .

the expression ((sqrt(1+sin x)+sqrt(1-sin x))/(sqrt(1+sin x)-sqrt(1-sin x)))=

(d)/(dx) [ 2 cot^(-1) ((sqrt(1+ sin x) + sqrt(1-sin x))/(sqrt(1+ sin x) - sqrt(1-sin x)))]=

Differentiate 'tan^(^^)(-1){(sqrt(1+sin x)+sqrt(1-sin x))/(sqrt(1+sin x)-sqrt(1-sin x))},darr backslash0

Prove that cot^(-1) ((sqrt(1 + sin x) + sqrt(1 - sin x))/(sqrt(1 + sin x) - sqrt(1 - sin x))) = (x)/(2), x in (0, (pi)/(4))