Find-Find-K-0-pi-2-tan-d-

Question Number 67463 by ~ À ® @ 237 ~ last updated on 27/Aug/19

F i n d

F i n d K = \int_{0}^{\frac{π}{2}} \sqrt{t a n θ} d θ

Commented by ~ À ® @ 237 ~ last updated on 27/Aug/19

U s i n g B (x, y) = 2 \int_{0_{}}^{\frac{π}{2}} {(s i n θ)}^{2 x - 1} {(c o s θ)}^{2 y - 1} d θ = \frac{Γ (x) Γ (y)}{Γ (x + y)}

2 K = 2 \int_{0}^{\frac{π}{2}} {(s i n θ)}^{\frac{1}{2}} {(c o s θ)}^{\frac{- 1}{2}} d θ = 2 \int_{0}^{\frac{π}{2}} {(s i n θ)}^{2 (\frac{3}{4}) - 1} {(c o s θ)}^{2 (\frac{1}{4}) - 1} d θ

= B (\frac{3}{4}, \frac{1}{4}) = \frac{Γ (\frac{3}{4}) Γ (\frac{1}{4})}{Γ (1)} = \frac{π}{s i n (\frac{3 π}{4})} = π \sqrt{2} u s i n g c o m p l e m e n t s f o r m u l a s Γ (x) Γ (1 - x) = \frac{π}{s i n (π x)}

S o K = \frac{π}{\sqrt{2}}

Commented by mathmax by abdo last updated on 27/Aug/19

l e t A = \int_{0}^{\frac{π}{2}} \sqrt{t a n θ} d θ c h a 7 g e m e n t \sqrt{t a n θ} = t g i v e t a n θ = t^{2} \Rightarrow

θ = a r c t a n (t^{2}) \Rightarrow A = \int_{0}^{\infty} \frac{t 2 t}{1 + t^{4}} d t

= 2 \int_{0}^{\infty} \frac{t^{2}}{1 + t^{4}} d t c h a n g e m e n t t = x^{\frac{1}{4}} g i v e A = 2 \int_{0}^{\infty} \frac{x^{\frac{1}{2}}}{1 + x} \times \frac{1}{4} x^{\frac{1}{4} - 1} d x

= \frac{1}{2} \int_{0}^{\infty} \frac{x^{\frac{1}{2} + \frac{1}{4} - 1}}{1 + x} d x = \frac{1}{2} \int_{0}^{\infty} \frac{x^{\frac{3}{4} - 1}}{1 + x} d x = \frac{1}{2} \times \frac{π}{s i n (\frac{3 π}{4})}

= \frac{π}{2 \times \frac{\sqrt{2}}{2}} = \frac{π}{\sqrt{2}} \Rightarrow ★ \int_{0}^{\frac{π}{2}} \sqrt{t a n θ} d θ = \frac{π}{\sqrt{2}} ★

i h a v e u s e d t h e r e s u l t \int_{0}^{\infty} \frac{t^{a - 1}}{1 + t} d t = \frac{π}{s i n (π a)} i f 0 < a < 1 .

Commented by ~ À ® @ 237 ~ last updated on 27/Aug/19

t h a n k s s i r

Answered by Kunal12588 last updated on 27/Aug/19

I = \int \sqrt{t a n x} d x

l e t \sqrt{t a n x} = t

\Rightarrow [\frac{1}{2 \sqrt{t a n x}} \times (1 + {t a n}^{2} x)] d x = d t

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

I = \int \frac{2 t^{2}}{1 + t^{4}} d t = \int \frac{1 + \frac{1}{t^{2}}}{t^{2} + \frac{1}{t^{2}}} d t + \int \frac{1 - \frac{1}{t^{2}}}{t^{2} + \frac{1}{t^{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}{\sqrt{2}} {t a n}^{- 1} (\frac{t^{2} - 1}{t \sqrt{2}}) + \frac{1}{2 \sqrt{2}} l o g ∣ \frac{t^{2} - \sqrt{2} t + 1}{t^{2} + \sqrt{2} t + 1} ∣ + C

= \frac{1}{\sqrt{2}} {t a n}^{- 1} (\frac{t a n x - 1}{\sqrt{2 t a n x}}) + \frac{1}{2 \sqrt{2}} l o g ∣ \frac{t a n x - \sqrt{2 t a n x} + 1}{t a n x + \sqrt{2 t a n x} + 1} ∣ + C

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

\underset{t \to \infty}{l i m} {\frac{1}{\sqrt{2}} {t a n}^{- 1} (\frac{t}{\sqrt{2}} - \frac{1}{t \sqrt{2}}) + \frac{1}{2 \sqrt{2}} l o g ∣ \frac{1 - \frac{\sqrt{2}}{t} + \frac{1}{t^{2}}}{1 + \frac{\sqrt{2}}{t} + \frac{1}{t^{2}}} ∣ + C}

= \frac{π}{2 \sqrt{2}}

\underset{t \to 0}{l i m} {\frac{1}{\sqrt{2}} {t a n}^{- 1} (\frac{t^{2} - 1}{t \sqrt{2}}) + \frac{1}{2 \sqrt{2}} l o g ∣ \frac{t^{2} - \sqrt{2} t + 1}{t^{2} + \sqrt{2} t + 1} ∣ + C}

= - \frac{π}{2 \sqrt{2}}

\int_{0}^{\frac{π}{2}} \sqrt{t a n x} d x = \frac{π}{2 \sqrt{2}} + \frac{π}{2 \sqrt{2}} = \frac{π}{\sqrt{2}} = \frac{π \sqrt{2}}{2}

Commented by ~ À ® @ 237 ~ last updated on 27/Aug/19

T h a n k s y o u s i r