tan-x-cot-x-dx-

Question Number 88339 by Rio Michael last updated on 10/Apr/20

\int (\sqrt{\tan x} + \sqrt{\cot x}) d x = ?

Commented by jagoll last updated on 10/Apr/20

\sqrt{\tan x} + \frac{1}{\sqrt{\tan x}} = \frac{\tan x + 1}{\sqrt{\tan x}}

\sqrt{\tan x} = t, \tan x = t^{2}

2 \sec^{2} x dx = 2 t dt

dx = \frac{t}{1 + t^{4}} dt]

\int \frac{t^{2} + 1}{t} \times \frac{t}{t^{4} + 1} dt = \int \frac{t^{2} + 1}{t^{4} + 1} dt

similar \int \frac{x^{2} + 1}{x^{4} + 1} dx

Commented by jagoll last updated on 10/Apr/20

\int \frac{t^{2} + 1}{t^{2} (t^{2} + \frac{1}{t^{2}})} dt = \int \frac{1 + \frac{1}{t^{2}}}{t^{2} + \frac{1}{t^{2}}} dt

= \int \frac{d (t - \frac{1}{t})}{{(t - \frac{1}{t})}^{2} + 2}

= \frac{1}{\sqrt{2}} arc \tan (\frac{t - \frac{1}{t}}{\sqrt{2}}) + c

= \frac{1}{\sqrt{2}} arc \tan (\frac{\sqrt{\tan x} - \frac{1}{\sqrt{\tan x}}}{\sqrt{2}}) + c

= \frac{1}{\sqrt{2}} arc \tan (\frac{\tan x - 1}{\sqrt{2 \tan x}}) + c

Commented by Rio Michael last updated on 10/Apr/20

sir, if \tan x = t^{2}

\Rightarrow \sec^{2} x = 2 t \frac{d t}{d x} \Rightarrow \sec^{2} x d x = 2 t d t

how did you get 2 \sec^{2} x d x = 2 t d t ? ?

Answered by TANMAY PANACEA. last updated on 10/Apr/20

m e t h o d - 1

\int \frac{\sqrt{s i n x}}{\sqrt{c o s x}} + \frac{\sqrt{c o s x}}{\sqrt{s i n x}} d x

\int \frac{s i n x + c o s x}{\frac{1}{\sqrt{2}} \times \sqrt{2 s i n x c o s x}} d x

\sqrt{2} \int \frac{d (s i n x - c o s x)}{\sqrt{1 - 1 + 2 s i n x c o s x}} d x

\sqrt{2} \int \frac{d (s i n x - c o s x)}{\sqrt{1 - {(s i n x - c o s x)}^{2}}} = \sqrt{2} {s i n}^{- 1} (\frac{s i n x - c o s x}{1}) + c

Answered by TANMAY PANACEA. last updated on 10/Apr/20

m e t h o d - 2

t^{2} = t a n x

2 t d t = {s e c}^{2} x d x

d x = \frac{2 t}{1 + t^{4}} d t

\int (t + \frac{1}{t}) \times \frac{2 t}{1 + t^{4}} d t

2 \int \frac{t (t + \frac{1}{t})}{t^{2} (t^{2} + \frac{1}{t^{2}})} d t

2 \int \frac{1 + \frac{1}{t^{2}}}{{(t - \frac{1}{t})}^{2} + 2} d t

2 \int \frac{d (t - \frac{1}{t})}{{(t - \frac{1}{t})}^{2} + 2} d t

2 \times \frac{1}{\sqrt{2}} {t a n}^{- 1} (\frac{t - \frac{1}{t}}{\sqrt{2}}) + c = \sqrt{2} {t a n}^{- 1} (\frac{\sqrt{t a n x} - \sqrt{c o t x}}{\sqrt{2}}) + c

Commented by M±th+et£s last updated on 10/Apr/20

n i c e s o l u t i o n s i r

Answered by M±th+et£s last updated on 10/Apr/20

= \int (\sqrt{t a n (x)} + \sqrt{c o t (x)}) \frac{t a n (x) + c o t (x)}{t a n (x) + c o t (x)} d x

= \int \frac{1}{t a n (x) + c o t (x)} (t a n (x) \sqrt{t a n (x)} + \frac{1}{\sqrt{t a n (x)}} + \frac{1}{\sqrt{c o t (x)}} + c o t (x) \sqrt{c o t (x)}) d x

= \int \frac{1}{t a n (x) + c o t (x)} (\frac{{t a n}^{2} (x) + 1}{\sqrt{t a n (x)}} + \frac{1 + {c o t}^{2} (x)}{\sqrt{c o t (x)}}) d x

= \int \frac{1}{2 + t a n (x) - 2 + c o t (x)} (\frac{{s e c}^{2} (x)}{\sqrt{t a n (x)}} + \frac{{c s c}^{2} (x)}{\sqrt{c o t (x)}}) d x

= \int \frac{6}{2 + {(\sqrt{t a n (x)} - \sqrt{c o t (x)})}^{2}} (\frac{{s e c}^{2} (x)}{2 \sqrt{t a n (x)}} - \frac{- {c s c}^{2} (x)}{2 \sqrt{c o t (x)}}) d x

= \sqrt{2} {t a n}^{- 1} (\frac{\sqrt{t a n (x)} - \sqrt{c o t (x)}}{2}) + c