Question

integration of root cosec X - 1

Answer 1

∫ √(1 + cosecx) dx

= ∫ √[(1 + cosecx)*(cosecx - 1)] / √(cosecx - 1) dx

= ∫ cotx / √(cosecx - 1) dx

= ∫ cosecx cotx dx / [cosecx √(cosecx - 1)]

Let cosecx - 1 = u^2

=> - cosecx cotx dx = 2udu

=> integral

= - 2 ∫ u du / [(u^2 + 1) * u]

= - 2 arctanu + c

= - 2arctan(√(cosecx - 1) + c.

hope it will help u

Answer 2

Answer:

$\int \sqrt(1 - cosecx) dx$

$= - \int \sqrt{(1 - cosecx)\times(cosecx + 1)} / \sqrt{(cosecx + 1) dx}$

$= \int {cotx} / \sqrt{(cosecx + 1) dx}$

$= \int cosecx. cotx. dx / {cosecx \sqrt{(cosecx + 1)} ------- (1)$

Let $cosecx +1 = u^2$

$= - cosecx.cotx.dx = 2u.du$

Substitute in the equation (1)

$= \int 2u / (-cosecx)\times(u).du$

$= -2 \int 1/ u2 -1 du$

By doing partial fractions:

$= -1/2 \int 1/1-u du + \frac{1}{2} \int 1/1+u du$

$= -\frac{1}{2} ln(1-u) + \frac{1}{2} ln(1+u) + C$

$= -\frac{1}{2} ln(1-\sqrt {cosecx +1}) + \frac{1}{2} ln (1+\sqrt {cosecx +1}) +C$