tan-x-dx-Who-can-solve-this-problem-

Question Number 93304 by Shakhzod last updated on 12/May/20

\int \sqrt{\tan (x)} d x W h o c a n s o l v e t h i s p r o b l e m ?

Commented by john santu last updated on 12/May/20

= - \frac{1}{\sqrt{2}} \tanh^{- 1} (\frac{\sqrt{\tan x} + \sqrt{\cot x}}{\sqrt{2}}) +

\frac{1}{\sqrt{2}} \tan^{- 1} (\frac{\sqrt{\tan x} - \sqrt{\cot x}}{\sqrt{2}}) + c

Commented by i jagooll last updated on 12/May/20

cooll man ��

Commented by i jagooll last updated on 12/May/20

set u = \sqrt{\tan x}, dx = \frac{2 u du}{u^{4} + 1}

\int \sqrt{\tan x} dx = \int \frac{2}{u^{2} + \frac{1}{u^{2}}} du

= \int \frac{1 - \frac{1}{u}}{{(u + \frac{1}{u})}^{2} - 2} du + \int \frac{1 + \frac{1}{u^{2}}}{{(u - \frac{1}{u})}^{2} + 2} du

[set u + \frac{1}{u} = \tan q], [w = u - \frac{1}{u}]

= \int \frac{1}{q^{2} - 2} dq + \int \frac{1}{w^{2} + 2} dw

= - \frac{1}{\sqrt{2}} \tanh^{- 1} (\frac{q}{\sqrt{2}}) + \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{w}{\sqrt{2}}) + c

= - \frac{1}{\sqrt{2}} \tanh^{- 1} (\frac{\sqrt{\tan x} + \sqrt{\cot x}}{\sqrt{2}})

+ \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{\sqrt{\tan x} - \sqrt{\cot x}}{\sqrt{2}}) + c

Commented by M±th+et+s last updated on 12/May/20

a t e n d p l s c h e c k f o r t y p o s a n d t h a n k y o u

Commented by mathmax by abdo last updated on 12/May/20

t h i s i n t e g r a l i s s o l v e d s e e t h e p l a t f o r m .

Answered by prakash jain last updated on 12/May/20

Commented by Shakhzod last updated on 12/May/20

G r e a t . T h a n k y o u s o m u c h .

Answered by M±th+et+s last updated on 12/May/20

= \frac{1}{\sqrt{2}} \int \frac{2 s i n (x)}{\sqrt{s i n (2 x)}}

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

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

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

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

i c a n s o l v e i t w i t h u s i n g a n o t h e r w a y s

b u t i d o n t h a v e a t i m e n o w i w i l l p o s t

t h e m i n a n o t h e t i m e

Commented by Shakhzod last updated on 12/May/20

T h a n k y o u . I w i l l w a i t y o u t h e n .

Answered by M±th+et+s last updated on 12/May/20

Answered by M±th+et+s last updated on 12/May/20

Answered by M±th+et+s last updated on 12/May/20

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

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

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

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

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

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