find-the-value-of-0-pi-2-tanx-dx-

Question Number 27828 by abdo imad last updated on 15/Jan/18

f i n d t h e v a l u e o f \int_{0}^{\frac{π}{2}} \sqrt{t a n x} d x .

Commented by NECx last updated on 15/Jan/18

t h a n k s f o r t h i s q u e s t i o n .

I r e a l l y n e e d t h e a n s w e r .

Commented by abdo imad last updated on 15/Jan/18

l e t u s e t h e c h . \sqrt{t a n x} = t \Leftrightarrow t a n x = t^{2} \Leftrightarrow x = a r c t a n (t^{2})

I = \int_{0}^{\frac{π}{2}} \sqrt{t a n x} d x = \int_{0}^{\infty} t \frac{2 t}{1 + t^{4}} d t = \int_{0}^{\propto} \frac{2 t^{2}}{1 + t^{4}} d t

= \frac{1}{2} \int_{- \propto}^{+ \propto} \frac{2 t^{2}}{1 + t^{4}} d t l e t i n t r o d u c e t h e c o m p l e x f u n c t i o n

f (z) = \frac{2 z^{2}}{1 + z^{4}} l e t f i n d t h e p o l e s o f f ?

1 + z^{4} = 0 \Leftrightarrow z^{4} = e^{i (2 k + 1) π} s o t h e p o l e s a r e z_{k} = e^{i (2 k + 1) \frac{π}{4}}

a n d k \in [[0, 3]] w e h a v e z_{0} = e^{i \frac{π}{4}}, z_{1} = e^{i \frac{3 π}{4}}, z_{2} = e^{i \frac{5 π}{4}}

z_{3} = e^{i \frac{7 π}{4}} a n d \int_{R^{}} f (z) d z = 2 i π (R e s (f, z_{0}) + R e s (f, z_{1}))

R e s (f, z_{0}) = \frac{2 z_{0}^{2}}{4 z_{0}^{3}} = \frac{1}{2} z_{0}^{- 1} = \frac{1}{2} e^{- i \frac{π}{4}}

R e s (f, z_{1}) = \frac{2 z_{1}^{2}}{4 z_{1}^{3}} = \frac{1}{2} z_{1}^{- 1} = - \frac{1}{2} e^{- i \frac{3 π}{4}} = \frac{1}{2} e^{- i (π - \frac{π}{4})}

= - \frac{1}{2} e^{i \frac{π}{4}}

\int_{R} f (z) d z = 2 i π [\frac{1}{2} (e^{- i \frac{π}{4}} - e^{i \frac{π}{4}})] = i π (- 2 i s i n (\frac{π}{4}))

= 2 π \frac{\sqrt{2}}{2} = π \sqrt{2} \Rightarrow I = \frac{1}{2} \int_{R} f (z) d z = \frac{π \sqrt{2}}{2} .

Answered by ajfour last updated on 15/Jan/18

l e t \sqrt{\tan x} = t

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

\int_{0}^{π / 2} \sqrt{\tan x} d x = \int_{0}^{\infty} \frac{2 t^{2} d t}{1 + t^{4}}

= \int_{0}^{\infty} \frac{2 d t}{(\frac{1}{t^{2}} + t^{2})}

= \int_{0}^{\infty} \frac{(1 - \frac{1}{t^{2}}) d t}{{(t + \frac{1}{t})}^{2} - 2} + \int_{0}^{\infty} \frac{(1 + \frac{1}{t^{2}}) d t}{{(t - \frac{1}{t})}^{2} + 2}

= \frac{1}{2 \sqrt{2}} \ln (\frac{t + \frac{1}{t} - \sqrt{2}}{t + \frac{1}{t} + \sqrt{2}}) ∣_{0}^{\infty} + \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{t - \frac{1}{t}}{\sqrt{2}}) ∣_{0}^{\infty}

= 0 + \frac{1}{\sqrt{2}} (\frac{π}{2} + \frac{π}{2}) = \frac{π}{\sqrt{2}} .

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