Question

Evaluate:

`int(sqrt(tanx) + sqrt(cotx))dx`

Answer 1

I = `int(sqrt(tanx) + 1/sqrt(tanx))dx`

= `int(tanx + 1)/sqrt(tanx)dx`

Put `sqrt(tanx)` = t

∴ tan x = t²

∴ x = tan^–1t²

∴ 1 dx = `1/(1 + (t^2)^2) 2t  dt`

∵ sec²x dx = 2t dt

∴ dx = `(2t)/(sec^2x)dx`

= `(2t)/(1 + tan^2x)dx`

= `(2t)/(1 + t^4)dt`

= `int(t^2 + 1)/t (2t)/(1 + t^4)dt`

= `2int(t^2 + 1)/(t^4 + 1)dt`

= `2int(1 + 1/t^2)/(t^2 + 1/t^2)dt`

= `2int((1 + 1/t^2))/((t - 1/t)^2 + 2)dt`

Put `t - 1/t` = u

∵ `[d/dt(t - 1/t) = 1 + 1/t^2]`

∴ `(t - (-1/t^2))dt` = 1 du

∴ `(1 + 1/t^2)dt` = 1 du

I = `2int1/(u^2 + 2)du`

= `2int1/(u^2 + (sqrt(2))^2)du`

= `2 1/sqrt(2) tan^-1(u/sqrt(2)) + c`

= `sqrt(2) tan^-1((t - 1/t)/sqrt(2)) + c`

= `sqrt(2) tan^-1 ((t^2 - 1)/(sqrt(2)t)) + c`

= `sqrt(2) tan^-1 ((tanx - 1)/(sqrt(2)*sqrt(tanx))) + c`