find-f-0-pi-4-1-tant-dt-with-gt-0-also-calculate-0-pi-4-tant-1-tant-dt-

Question Number 51551 by maxmathsup by imad last updated on 28/Dec/18

f i n d f (λ) = \int_{0}^{\frac{π}{4}} \sqrt{1 + λ t a n t} d t w i t h λ > 0

a l s o c a l c u l a t e \int_{0}^{\frac{π}{4}} \frac{t a n t}{\sqrt{1 + λ t a n t}} d t .

Commented by Abdo msup. last updated on 29/Dec/18

c h a n g e m e n t t a n t = x g i v e

f (λ) = \int_{0}^{1} \sqrt{1 + λ x} \frac{d x}{1 + x^{2}} = \int_{0}^{1} \frac{\sqrt{1 + λ x}}{1 + x^{2}} d x

=_{λ x = {s h}^{2} u} \int_{0}^{a r g s h (\sqrt{λ})} \frac{c h u}{1 + \frac{1}{λ^{2}} {s h}^{4} u} \frac{1}{λ} 2 s h u c h u d u

= \frac{2}{λ} \int_{0}^{l n (\sqrt{λ} + \sqrt{1 + λ})} \frac{s h (u) {c h}^{2} (u)}{λ^{2} + {s h}^{4} u} λ^{2} d u

= 2 λ \int_{0}^{l n (\sqrt{λ} + \sqrt{1 + λ})} \frac{s h (u) \frac{1 + c h (2 u)}{2}}{λ^{2} + {(\frac{c h (2 u) - 1}{2})}^{2}} d u

= 4 λ \int_{0}^{l n (\sqrt{λ} + \sqrt{1 + λ})} \frac{(1 + c h (2 u)) s h (u)}{4 λ^{2} + {(c h (2 u) - 1)}^{2}} d u

= 4 λ \int_{0}^{l n (\sqrt{λ} + \sqrt{1 + λ})} \frac{(1 + \frac{e^{2 u} + e^{- 2 u}}{2}) (\frac{e^{u} - e^{- u}}{2})}{4 λ^{2} + {(\frac{e^{2 u} + e^{- 2 u}}{2} - 1)}^{2}} d u

= 4 λ \int_{0}^{l n (\sqrt{λ} + \sqrt{1 + λ})} \frac{(2 + e^{2 u} + e^{- 2 u}) (e^{u} - e^{- u})}{16 λ^{2} + {(e^{2 u} + e^{- 2 u} - 2)}^{2}} d u

=_{e^{u} = t} 4 λ \int_{1}^{\sqrt{λ} + \sqrt{1 + λ}} \frac{(2 + t^{2} + t^{- 2}) (t - t^{- 1})}{16 λ^{2} + {(t^{2} + t^{- 2} - 2)}^{2}} \frac{d t}{t}

= 4 λ \int_{1}^{\sqrt{λ} + \sqrt{1 + λ}} \frac{(t^{4} + 2 t^{2} + 1) (t^{2} - 1)}{t^{4} (16 λ^{2} + {(t^{2} + t^{- 2} - 2)}^{2})} d t

\dots b e c o n t i n u e d \dots

Commented by Abdo msup. last updated on 29/Dec/18

l e t u s e t h e c h a n g e m e n t \sqrt{1 + λ t a n t} = u \Rightarrow

1 + λ t a n t = u^{2} \Rightarrow λ t a n t = u^{2} - 1 \Rightarrow t a n t = \frac{u^{2} - 1}{λ} \Rightarrow

t = a r c t a n (\frac{1}{λ} u^{2} - \frac{1}{λ}) \Rightarrow

f (λ) = \int_{1}^{\sqrt{1 + λ}} u . \frac{2}{λ} \frac{u}{1 + {(\frac{1}{λ} u^{2} - \frac{1}{λ})}^{2}} d u

= \frac{2}{λ} \int_{1}^{\sqrt{1 + λ}} \frac{u^{2}}{λ^{2} + {(u^{2} - 1)}^{2}} d u

=_{u = c h (t)} \frac{2}{λ} \int_{a r g c h (1)}^{a r g c h (\sqrt{1 + λ})} \frac{{c h}^{2} t}{λ^{2} + {s h}^{4} t} s h (t) d t

\dots b e c o n t i n u e d \dots

Answered by tanmay.chaudhury50@gmail.com last updated on 29/Dec/18

g (t) = \sqrt{1 + λ t a n t}

g (0) = 1

g (\frac{π}{4}) = \sqrt{1 + λ}

\int_{0}^{\frac{π}{4}} \sqrt{1 + λ t a n t} d t \approx \frac{1}{2} (1 + \sqrt{1 + λ}) \times \frac{π}{4}

\int_{0}^{\frac{π}{4}} \sqrt{1 + λ t a n t} d t \approx \frac{π}{8} (1 + \sqrt{1 + λ}) [a p p r o x m i t a t i o n]

Answered by tanmay.chaudhury50@gmail.com last updated on 29/Dec/18

g (t) = \sqrt{1 + λ t a n t}

g (0) = 1

g (\frac{π}{4}) = \sqrt{1 + λ}

\int_{0}^{\frac{π}{4}} \sqrt{1 + λ t a n t} d t \approx \frac{1}{2} (1 + \sqrt{1 + λ}) \times \frac{π}{4}

\int_{0}^{\frac{π}{4}} \sqrt{1 + λ t a n t} d t \approx \frac{π}{8} (1 + \sqrt{1 + λ}) [a p p r o x m i t a t i o n]