evaluate-tan-d-

Question Number 43490 by peter frank last updated on 11/Sep/18

evaluate

\int \sqrt{\tan θ d θ}

Commented by maxmathsup by imad last updated on 11/Sep/18

l e t I = \int \sqrt{t a n θ} d θ c h a n g e m e n t \sqrt{t a n θ} = x g i v e t a n θ = x^{2} ⇎ θ = a r c t a n (x^{2}) \Rightarrow

I = \int x \frac{2 x}{1 + x^{4}} d x = \int \frac{2 x^{2}}{1 + x^{4}} d x l e t d e c o m p o s e F (x) = \frac{2 x^{2}}{1 + x^{4}}

F (x) = \frac{2 x^{2}}{{(x^{2} + 1)}^{2} - 2 x^{2}} = \frac{2 x^{2}}{(x^{2} + \sqrt{2} x + 1) (x^{2} - \sqrt{2} x + 1)}

= \frac{a x + b}{x^{2} + \sqrt{2} x + 1} + \frac{c x + d}{x^{2} - \sqrt{2} x + 1}

F (- x) = F (x) \Rightarrow \frac{- a x + b}{x^{2} - \sqrt{2} x + 1} + \frac{- c x + d}{x^{2} + \sqrt{2} x + 1} = F (x) \Rightarrow c = - a a n d b = d \Rightarrow

F (x) = \frac{a x + b}{x^{2} + \sqrt{2} x + 1} + \frac{- a x + b}{x^{2} - \sqrt{2} x + 1}

F (0) = 0 = 2 b \Rightarrow b = 0 \Rightarrow F (x) = \frac{a x}{x^{2} + \sqrt{2} x + 1} - \frac{a x}{x^{2} - \sqrt{2} x + 1}

F (1) = 1 = \frac{a}{2 + \sqrt{2}} - \frac{a}{2 - \sqrt{2}} \Rightarrow (\frac{- 2 \sqrt{2}}{2}) a = 1 \Rightarrow a = - \frac{1}{\sqrt{2}} \Rightarrow

F (x) = \frac{1}{\sqrt{2}} {\frac{1}{x^{2} - \sqrt{2} x + 1} - \frac{1}{x^{2} + \sqrt{2} x + 1}} \Rightarrow

I = \frac{1}{\sqrt{2}} \int \frac{d x}{x^{2} - \sqrt{2} x + 1} - \frac{1}{\sqrt{2}} \int \frac{d x}{x^{2} + \sqrt{2} x + 1} b u t

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

=_{x - \frac{1}{\sqrt{2}} = \frac{u}{\sqrt{2}}} \int \frac{1}{\frac{1}{2} (1 + u^{2})} \frac{d u}{\sqrt{2}} = \sqrt{2} \int \frac{d u}{1 + u^{2}} = \sqrt{2} a r c t a n (u) + c_{1}

= \sqrt{2} a r c t a n (x \sqrt{2} - 1) + c_{1} = \sqrt{2} a r c t a n (\sqrt{2 t a n θ} - 1) + c_{1} a l s o w e h a v e

\int \frac{d x}{x^{2} + \sqrt{2} x + 1} = \int \frac{d x}{{(x + \frac{1}{\sqrt{2}})}^{2} + \frac{1}{2}} =_{x + \frac{1}{\sqrt{2}} = \frac{u}{\sqrt{2}}} \int \frac{1}{\frac{1}{2} (1 + u^{2})} \frac{d u}{\sqrt{2}}

= \sqrt{2} a r c t a n (\sqrt{2} x + 1) = \sqrt{2} a r c t a n (\sqrt{2 t a n θ} + 1) \Rightarrow

\int \sqrt{t a n θ} d θ = a r c t a n (\sqrt{2 t a n θ} - 1) - a r c t a n (\sqrt{2 t a n θ} + 1) + c

Commented by peter frank last updated on 11/Sep/18

t h a n k s

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Sep/18

t^{2} = t a n θ

2 t d t = {s e c}^{2} θ d θ

\frac{2 t}{1 + t^{4}} d t = d θ

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

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

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

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

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

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

\frac{1}{2 \sqrt{2}} l n ∣ \frac{(t + \frac{1}{t}) - \sqrt{2}}{(t + \frac{1}{t}) + \sqrt{2}} ∣ + \frac{1}{\sqrt{2}} {t a n}^{- 1} (\frac{t - \frac{1}{t}}{\sqrt{2}}) + c

\frac{1}{2 \sqrt{2}} l n ∣ \frac{\sqrt{t a n θ} + \frac{1}{\sqrt{t a n θ}} - \sqrt{2}}{\sqrt{t a n θ} + \frac{1}{\sqrt{t a n θ}} + \sqrt{2}} ∣ + \frac{1}{\sqrt{2}} {t a n}^{- 1} (\frac{\sqrt{t a n θ} - \frac{1}{\sqrt{t a n θ}}}{\sqrt{2}}) + c

Commented by peter frank last updated on 11/Sep/18

t h a n k s