tanx-cotxdx-

Question Number 76298 by benjo last updated on 26/Dec/19

\int \sqrt{tanx + cotxdx}

Commented by benjo last updated on 26/Dec/19

Commented by john santuy last updated on 26/Dec/19

u s i n g W e i r s t r r a s m e t o d e .

l e t t = t a n (\frac{x}{2})

Commented by benjo last updated on 26/Dec/19

please sir your write step by step

Commented by mathmax by abdo last updated on 26/Dec/19

l e t I = \int \sqrt{t a n x + \frac{1}{t a n x}} d x \Rightarrow I = \int \sqrt{\frac{s i n x}{c o s x} + \frac{c o s x}{s i n x}} d x

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

v h a n g e m e n t \sqrt{s i n (2 x)} = t g i v e I = \int \frac{\sqrt{2}}{t} \frac{t d t}{\sqrt{1 - t^{4}}} b e c a u s e

s i n (2 x) = t^{2} \Rightarrow 2 x = a r c s i n (t^{2}) \Rightarrow 2 d x = \frac{2 t d t}{\sqrt{1 - t^{4}}}

\Rightarrow I = \int \frac{\sqrt{2} d t}{\sqrt{1 - t^{4}}} a n d t h i s i n t e g r a l i s n o t r e s o l u b l e b y e l e m e n t a r y

f u n c t i o n s . .

Answered by john santuy last updated on 26/Dec/19

I = \int s e c x \sqrt{c o t x} d x

l e t c o t x = u^{2} \to - {c s c}^{2} x d x = 2 u d u

d x = - \frac{2 u d u}{\sqrt{1 + u^{4}}}

I = \int \frac{\sqrt{1 + u^{4}}}{u^{2}} \times u^{2} \times \frac{2 u}{\sqrt{1 + u^{4}}} d u

= - 2 \int u d u = - u^{2} + C

h e n c e - c o t x + C

Commented by benjo last updated on 26/Dec/19

waw thanks sir

Commented by MJS last updated on 26/Dec/19

sorry but {that}^{'} s wrong

\int \sqrt{\tan x + \cot x} d x = \int \frac{d x}{\sqrt{\sin x \cos x}} =

= \sqrt{2} \int \frac{d x}{\sqrt{\sin 2 x}}

and this cannot be solved using elementary

calculus

Commented by MJS last updated on 26/Dec/19

but \frac{d}{d x} [- \cot x] = \csc^{2} x \neq \sec x \sqrt{\cot x}

Commented by john santu last updated on 26/Dec/19

tanx + \frac{1}{tanx} = \frac{\tan^{2} x + 1}{tanx} = \frac{\sec^{2} x}{tanx}

\sqrt{\frac{\sec^{2} x}{tanx}} = secx \sqrt{cotx} sir ?

Commented by john santu last updated on 26/Dec/19

o o y e s s i r . i^{'} m w r o n g i n d x = - \frac{2 u}{{(\sqrt{1 + u^{2}})}^{2}} d u