1-find-the-value-of-pi-4-pi-3-1-tan-d-

Question Number 43535 by maxmathsup by imad last updated on 11/Sep/18

1) find the value of ∫_(π/4) ^(π/3) (√(1+tanθ))dθ .

1) f i n d t h e v a l u e o f \int_{\frac{π}{4}}^{\frac{π}{3}} \sqrt{1 + t a n θ} d θ .

Commented by MJS last updated on 12/Sep/18

it′s possible but it′s a long way start with (√(1+tan θ))=sec^2 θ ((√(1+tan θ))/(1+tan^2 θ)) then t=1+tan θ → dθ=(dt/(sec^2 θ)) which leads to ((√t)/(1+(t−1)^2 )) then u=(√t) → dt=2(√t)du which leads to ((2v^2 )/(1+(v^2 −1)^2 ))=((2v^2 )/(v^4 −2v^2 +2)) then factorise the denominator and decompose the fraction...

{it}^{'} s possible but {it}^{'} s a long way

start with

\sqrt{1 + \tan θ} = \sec^{2} θ \frac{\sqrt{1 + \tan θ}}{1 + \tan^{2} θ}

then t = 1 + \tan θ \to d θ = \frac{d t}{\sec^{2} θ} which leads

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

then u = \sqrt{t} \to d t = 2 \sqrt{t} d u which leads to

\frac{2 v^{2}}{1 + {(v^{2} - 1)}^{2}} = \frac{2 v^{2}}{v^{4} - 2 v^{2} + 2}

then factorise the denominator and decompose

the fraction \dots

Commented by maxmathsup by imad last updated on 12/Sep/18

let A = ∫_(π/4) ^(π/3) (√(1+tanθ))dθ changement (√(1+tanθ))=x give 1+tanθ =x^2 ⇒ θ =artan(x^2 −1) ⇒dθ =((2x)/(1+(x^2 −1)^2 ))dx ⇒ A = ∫_(√2) ^(√(1+(√3))) x ((2xdx)/(1+(x^2 −1)^2 )) = ∫_(√2) ^(√(1+(√3))) ((2x^2 )/(1+x^4 −2x^2 +1))dx =∫_(√2) ^(√(1+(√3))) ((2x^2 )/(x^4 −2x^2 +2))dx let decompose inside C(x) F(x)=((2x^2 )/(x^4 −2x^2 +2)) roots of x^4 −2x^2 +2 →x^2 =t ⇒t^2 −2t +2 →Δ^′ =−1 ⇒ z_1 =1+i and z_2 =1−i ⇒F(x)=(a/(x−z_1 )) +(b/(x−z_2 )) a=lim_(x→z_1 ) (x−z_1 )F(x)=((2z_1 ^2 )/(z_1 −z_2 )) = ((2(1+i)^2 )/(2i)) =2 b =lim_(x→z_2 ) (x−z_2 )F(x)=((2z_2 ^2 )/(z_2 −z_1 )) = ((2(1−i)^2 )/(−2i)) =2 ⇒ F(x) =(2/(x^2 −1−i)) +(2/(x^2 −1+i)) ⇒ A = ∫_(√2) ^(√(1+(√3))) ((2dx)/(x^2 −1−i)) + ∫_(√2) ^(√(1+(√3))) ((2dx)/(x^2 −1+i)) =H +H^− = 2Re(H) = 2 Re( ∫_(√2) ^(√(1+(√3))) ((2dx)/(x^2 −1−i))) let decompose f(x) =(2/(x^2 −(1+i))) ⇒f(x)= (2/((x−(√(1+i)))(x+(√(1+i))))) wd have 1+i =(√2)e^((iπ)/4) ⇒(√(1+i))=2^(1/4) e^((iπ)/8) ⇒f(x) =(2/((x−2^(1/4) e^((iπ)/8) )(x+2^(1/4) e^((iπ)/8) ))) =(a/(x−2^(1/4) e^((iπ)/8) )) +(b/(x+2^(1/4) e^((iπ)/8) )) a = (2/(2 2^(1/4) e^((iπ)/8) )) =2^(−(1/4)) e^(−((iπ)/8)) and b = (2/(−2.2^(1/4) e^((iπ)/8) )) =−2^(1/4) e^(−((iπ)/8)) ⇒ f(x) =2^(1/4) e^(−((iπ)/8)) { (1/(x−2^(1/4) e^((iπ)/8) )) − (1/(x+2^(1/4) e^((iπ)/8) ))}⇒∫_(√2) ^(√(1+(√3))) ((2dx)/(x^2 −1−i)) =2^(1/4) e^(−((iπ)/8)) { ∫_(√2) ^(√(1+(√3))) (dx/(x−2^(1/4) e^((iπ)/8) )) −∫_(√2) ^(√(1+(√3))) (dx/(x+2^(1/4) e^((iπ)/8) ))} let find ∫ (dx/(x−z)) with z =a+ib ∫ (dx/(x−z)) = ∫ (dx/(x−a−ib)) = ∫ ((x−a +ib)/((x−a)^2 +b^2 )) dx =(1/2)ln{(x−a)^2 +b^2 } +ib ∫ (dx/((x−a)^2 +b^2 )) but ∫ (dx/((x−a)^2 +b^2 )) =_(x−a=bt) ∫ ((bdt)/(b^2 (1+t^2 ))) =(1/b) arctan(((x−a)/b)) ⇒ ∫ (dx/(x−z)) =(1/2)ln{(x−a)^2 +b^2 } +i arctan(((x−a)/b))we have 2^(1/4) e^((iπ)/8) =2^(1/4) cos((π/8)) +i 2^(1/4) sin((π/8)) =a+ib ⇒ ∫_(√2) ^(√(1+(√3))) (dx/(x−2^(1/4) e^((iπ)/8) )) =[(1/2)ln{(x−2^(1/4) cos((π/8)))^2 +(√2)sin^2 ((π/8))}]_(√2) ^(√(1+(√3))) +i[ arctan{((x−2^(1/4) cos((π/8)))/(2^(1/4) sin((π/8))))}]_(√2) ^(√(1+(√3))) so the value of A is known after calculus..

l e t A = \int_{\frac{π}{4}}^{\frac{π}{3}} \sqrt{1 + t a n θ} d θ c h a n g e m e n t \sqrt{1 + t a n θ} = x g i v e 1 + t a n θ = x^{2} \Rightarrow

θ = a r t a n (x^{2} - 1) \Rightarrow d θ = \frac{2 x}{1 + {(x^{2} - 1)}^{2}} d x \Rightarrow

A = \int_{\sqrt{2}}^{\sqrt{1 + \sqrt{3}}} x \frac{2 x d x}{1 + {(x^{2} - 1)}^{2}} = \int_{\sqrt{2}}^{\sqrt{1 + \sqrt{3}}} \frac{2 x^{2}}{1 + x^{4} - 2 x^{2} + 1} d x

= \int_{\sqrt{2}}^{\sqrt{1 + \sqrt{3}}} \frac{2 x^{2}}{x^{4} - 2 x^{2} + 2} d x l e t d e c o m p o s e i n s i d e C (x) F (x) = \frac{2 x^{2}}{x^{4} - 2 x^{2} + 2}

r o o t s o f x^{4} - 2 x^{2} + 2 \to x^{2} = t \Rightarrow t^{2} - 2 t + 2 \to Δ^{^{'}} = - 1 \Rightarrow

z_{1} = 1 + i a n d z_{2} = 1 - i \Rightarrow F (x) = \frac{a}{x - z_{1}} + \frac{b}{x - z_{2}}

a = {l i m}_{x \to z_{1}} (x - z_{1}) F (x) = \frac{2 z_{1}^{2}}{z_{1} - z_{2}} = \frac{2 {(1 + i)}^{2}}{2 i} = 2

b = {l i m}_{x \to z_{2}} (x - z_{2}) F (x) = \frac{2 z_{2}^{2}}{z_{2} - z_{1}} = \frac{2 {(1 - i)}^{2}}{- 2 i} = 2 \Rightarrow

F (x) = \frac{2}{x^{2} - 1 - i} + \frac{2}{x^{2} - 1 + i} \Rightarrow A = \int_{\sqrt{2}}^{\sqrt{1 + \sqrt{3}}} \frac{2 d x}{x^{2} - 1 - i} + \int_{\sqrt{2}}^{\sqrt{1 + \sqrt{3}}} \frac{2 d x}{x^{2} - 1 + i}

= H + \bar{H} = 2 R e (H) = 2 R e (\int_{\sqrt{2}}^{\sqrt{1 + \sqrt{3}}} \frac{2 d x}{x^{2} - 1 - i}) l e t d e c o m p o s e

f (x) = \frac{2}{x^{2} - (1 + i)} \Rightarrow f (x) = \frac{2}{(x - \sqrt{1 + i}) (x + \sqrt{1 + i})} w d h a v e

1 + i = \sqrt{2} e^{\frac{i π}{4}} \Rightarrow \sqrt{1 + i} = 2^{\frac{1}{4}} e^{\frac{i π}{8}} \Rightarrow f (x) = \frac{2}{(x - 2^{\frac{1}{4}} e^{\frac{i π}{8}}) (x + 2^{\frac{1}{4}} e^{\frac{i π}{8}})}

= \frac{a}{x - 2^{\frac{1}{4}} e^{\frac{i π}{8}}} + \frac{b}{x + 2^{\frac{1}{4}} e^{\frac{i π}{8}}}

a = \frac{2}{2 2^{\frac{1}{4}} e^{\frac{i π}{8}}} = 2^{- \frac{1}{4}} e^{- \frac{i π}{8}} a n d b = \frac{2}{- 2 . 2^{\frac{1}{4}} e^{\frac{i π}{8}}} = - 2^{\frac{1}{4}} e^{- \frac{i π}{8}} \Rightarrow

f (x) = 2^{\frac{1}{4}} e^{- \frac{i π}{8}} {\frac{1}{x - 2^{\frac{1}{4}} e^{\frac{i π}{8}}} - \frac{1}{x + 2^{\frac{1}{4}} e^{\frac{i π}{8}}}} \Rightarrow \int_{\sqrt{2}}^{\sqrt{1 + \sqrt{3}}} \frac{2 d x}{x^{2} - 1 - i}

= 2^{\frac{1}{4}} e^{- \frac{i π}{8}} {\int_{\sqrt{2}}^{\sqrt{1 + \sqrt{3}}} \frac{d x}{x - 2^{\frac{1}{4}} e^{\frac{i π}{8}}} - \int_{\sqrt{2}}^{\sqrt{1 + \sqrt{3}}} \frac{d x}{x + 2^{\frac{1}{4}} e^{\frac{i π}{8}}}} l e t f i n d

\int \frac{d x}{x - z} w i t h z = a + i b

\int \frac{d x}{x - z} = \int \frac{d x}{x - a - i b} = \int \frac{x - a + i b}{{(x - a)}^{2} + b^{2}} d x

= \frac{1}{2} l n {{(x - a)}^{2} + b^{2}} + i b \int \frac{d x}{{(x - a)}^{2} + b^{2}} b u t

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

\int \frac{d x}{x - z} = \frac{1}{2} l n {{(x - a)}^{2} + b^{2}} + i a r c t a n (\frac{x - a}{b}) w e h a v e

2^{\frac{1}{4}} e^{\frac{i π}{8}} = 2^{\frac{1}{4}} c o s (\frac{π}{8}) + i 2^{\frac{1}{4}} s i n (\frac{π}{8}) = a + i b \Rightarrow

\int_{\sqrt{2}}^{\sqrt{1 + \sqrt{3}}} \frac{d x}{x - 2^{\frac{1}{4}} e^{\frac{i π}{8}}} = {[\frac{1}{2} l n {{(x - 2^{\frac{1}{4}} c o s (\frac{π}{8}))}^{2} + \sqrt{2} {s i n}^{2} (\frac{π}{8})}]}_{\sqrt{2}}^{\sqrt{1 + \sqrt{3}}}

+ i {[a r c t a n {\frac{x - 2^{\frac{1}{4}} c o s (\frac{π}{8})}{2^{\frac{1}{4}} s i n (\frac{π}{8})}}]}_{\sqrt{2}}^{\sqrt{1 + \sqrt{3}}} s o t h e v a l u e o f A i s k n o w n a f t e r

c a l c u l u s . .

Answered by behi83417@gmail.com last updated on 12/Sep/18

tgθ=tg^2 x⇒(1+tg^2 θ)dθ=2tgx(1+tg^2 x)dx I=∫ (1/(cosx)).((2tgx(1+tg^2 x))/((1+tg^4 x)))dx= =∫ ((((2sinx)/(cosx)).(1/(cos^2 x)))/(cosx(((sin^4 x+cos^4 x)/(cos^4 x)))))dx= =∫ ((2sinx)/(sin^4 x+cos^4 x))dx= =∫((2sinx)/(1−2cos^2 x(1−cos^2 x)))dx= =∫((2sinx)/(2cos^4 x−2cos^2 x+1))dx=−∫(du/(2u^4 −2u^2 +1))... 2u^4 −2u^2 +1=0⇒u^2 =((2±(√(4−8)))/4)= =(1/2)(1±i)=α^2 ,β^2 (α,β∈C) [α^2 +β^2 =1,α^2 −β^2 =i,αβ=(1/( (√2)))] (α+β)^2 =1+2×(1/( (√2)))=(√2)+1⇒α+β=(√((√2)+1)) α−β=−((√2)−1)⇒α−β=i(√((√2)−1)) ⇒ { ((α=(1/2)[(√((√2)+1))+i(√((√2)−1))))),((β=(1/2)[(√((√2)+1))−i(√((√2)−1))))) :} I=∫(du/((u−α)(u+α)(u−β)(u+β)))= =(i/(2α)).ln((u+α)/(u−α))−(i/(2β)).ln((u+β)/(u−β))+const.= =(i/(2α))ln((u+α)/(u−β))−(i/(2β))ln((u+β)/(u−β))+const.== =(i/(2α))ln(((√(2+tgθ))+α)/( (√(2+tgθ))−α))−(i/(2β))ln(((√(2+tgθ))+β)/( (√(2+tgθ))−β))+const [(1/(Π(u−α)))=Σ(k/((u−α))) u=α⇒k_1 =(1/((2α)(α^2 −β^2 )))=((−i)/(2α)) u=−α⇒k_2 =(1/((2α)(β^2 −α^2 )))=(i/(2α)) u=β⇒k_3 =(1/(2β(β^2 −α^2 )))=(i/(2β)) u=−β⇒k_4 =(1/(2β(α^2 −β^2 )))=((−i)/(2β))]

t g θ = {t g}^{2} x \Rightarrow (1 + {t g}^{2} θ) d θ = 2 t g x (1 + {t g}^{2} x) d x

I = \int \frac{1}{c o s x} . \frac{2 t g x (1 + {t g}^{2} x)}{(1 + {t g}^{4} x)} d x =

= \int \frac{\frac{2 s i n x}{c o s x} . \frac{1}{{c o s}^{2} x}}{c o s x (\frac{{s i n}^{4} x + {c o s}^{4} x}{{c o s}^{4} x})} d x =

= \int \frac{2 s i n x}{{s i n}^{4} x + {c o s}^{4} x} d x =

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

= \int \frac{2 s i n x}{2 {c o s}^{4} x - 2 {c o s}^{2} x + 1} d x = - \int \frac{d u}{2 u^{4} - 2 u^{2} + 1} \dots

2 u^{4} - 2 u^{2} + 1 = 0 \Rightarrow u^{2} = \frac{2 \pm \sqrt{4 - 8}}{4} =

= \frac{1}{2} (1 \pm i) = α^{2}, β^{2} (α, β \in C)

[α^{2} + β^{2} = 1, α^{2} - β^{2} = i, α β = \frac{1}{\sqrt{2}}]

{(α + β)}^{2} = 1 + 2 \times \frac{1}{\sqrt{2}} = \sqrt{2} + 1 \Rightarrow α + β = \sqrt{\sqrt{2} + 1}

α - β = - (\sqrt{2} - 1) \Rightarrow α - β = i \sqrt{\sqrt{2} - 1}

\Rightarrow {\begin{cases} α = \frac{1}{2} [\sqrt{\sqrt{2} + 1} + i \sqrt{\sqrt{2} - 1}) \\ β = \frac{1}{2} [\sqrt{\sqrt{2} + 1} - i \sqrt{\sqrt{2} - 1}) \end{cases}

I = \int \frac{d u}{(u - α) (u + α) (u - β) (u + β)} =

= \frac{i}{2 α} . l n \frac{u + α}{u - α} - \frac{i}{2 β} . l n \frac{u + β}{u - β} + c o n s t . =

= \frac{i}{2 α} l n \frac{u + α}{u - β} - \frac{i}{2 β} l n \frac{u + β}{u - β} + c o n s t . ==

= \frac{i}{2 α} l n \frac{\sqrt{2 + t g θ} + α}{\sqrt{2 + t g θ} - α} - \frac{i}{2 β} l n \frac{\sqrt{2 + t g θ} + β}{\sqrt{2 + t g θ} - β} + c o n s t

[\frac{1}{Π (u - α)} = Σ \frac{k}{(u - α)}

u = α \Rightarrow k_{1} = \frac{1}{(2 α) (α^{2} - β^{2})} = \frac{- i}{2 α}

u = - α \Rightarrow k_{2} = \frac{1}{(2 α) (β^{2} - α^{2})} = \frac{i}{2 α}

u = β \Rightarrow k_{3} = \frac{1}{2 β (β^{2} - α^{2})} = \frac{i}{2 β}

u = - β \Rightarrow k_{4} = \frac{1}{2 β (α^{2} - β^{2})} = \frac{- i}{2 β}]

Answered by MJS last updated on 13/Sep/18

∫(√(1+tan θ)) dθ=∫((sec^2 θ)/(1+tan^2 θ))(√(1+tan θ)) dθ= [t=1+tan θ → dθ=(dt/(sec^2 θ))] =∫((√t)/(1+(t−1)^2 ))dt= [u=(√t) → dt=2(√t)du] =2∫(u^2 /(1+(u^2 −1)^2 ))du=2∫(u^2 /(u^4 −2u^2 +2))du= =2∫(u^2 /((u^2 −(√(2+(√8)))u+(√2))(u^2 +(√(2+(√8)))u+(√2))))du= [a=(√(2+(√8))); b=(√2)] =2∫(u^2 /((u^2 −au+b)(u^2 +au+b)))du= =2∫((u/(2a(u^2 −au+b)))−(u/(2a(u^2 +au+b))))du= =(1/a)∫(u/(u^2 −au+b))du−(1/a)∫(u/(u^2 +au+b))du ∫(u/(u^2 −au+b))du=∫(((2u−a)/(2(u^2 −au+b)))+(a/(2(u^2 −au+b)))du= =(1/2)∫((2u−a)/(u^2 −au+b))du+(a/2)∫(du/(u^2 −au+b))= [these are standard integrals] =(1/2)ln (u^2 −au+b) +(a/( (√(4b−a^2 ))))arctan ((2u−a)/( (√(4b−a^2 )))) ∫(u/(u^2 +au+b))du=∫(((2u+a)/(2(u^2 +au+b)))−(a/(2(u^2 +au+b))))du= =(1/2)∫((2u+a)/(u^2 +au+b))du−(a/2)∫(du/(u^2 +au+b))= [again standard integrals] =(1/2)ln (u^2 +au+b) −(a/( (√(4b−a^2 ))))arctan ((2u+a)/( (√(4b−a^2 )))) =(1/(2a))(ln (u^2 −au+b) +ln (u^2 +au+b))+(1/( (√(4b−a^2 ))))(arctan ((2u−a)/( (√(4b−a^2 )))) −arctan ((2u+a)/( (√(4b−a^2 )))))= =(1/(2a))ln (u^4 +(2b−a^2 )u^2 +b^2 ) +(1/( (√(4b−a^2 ))))(arctan ((2u−a)/( (√(4b−a^2 )))) −arctan ((2u+a)/( (√(4b−a^2 )))))= =(1/(2a))ln ∣t^2 +(2b−a^2 )t+b^2 ∣ +(1/( (√(4b−a^2 ))))(arctan ((2(√t)−a)/( (√(4b−a^2 )))) −arctan ((2(√t)+a)/( (√(4b−a^2 )))))+C sorry no time to finish t=1+tan θ a=(√(2+(√8))) b=(√2)

\int \sqrt{1 + \tan θ} d θ = \int \frac{\sec^{2} θ}{1 + \tan^{2} θ} \sqrt{1 + \tan θ} d θ =

[t = 1 + \tan θ \to d θ = \frac{d t}{\sec^{2} θ}]

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

[u = \sqrt{t} \to d t = 2 \sqrt{t} d u]

= 2 \int \frac{u^{2}}{1 + {(u^{2} - 1)}^{2}} d u = 2 \int \frac{u^{2}}{u^{4} - 2 u^{2} + 2} d u =

= 2 \int \frac{u^{2}}{(u^{2} - \sqrt{2 + \sqrt{8}} u + \sqrt{2}) (u^{2} + \sqrt{2 + \sqrt{8}} u + \sqrt{2})} d u =

[a = \sqrt{2 + \sqrt{8}}; b = \sqrt{2}]

= 2 \int \frac{u^{2}}{(u^{2} - a u + b) (u^{2} + a u + b)} d u =

= 2 \int (\frac{u}{2 a (u^{2} - a u + b)} - \frac{u}{2 a (u^{2} + a u + b)}) d u =

= \frac{1}{a} \int \frac{u}{u^{2} - a u + b} d u - \frac{1}{a} \int \frac{u}{u^{2} + a u + b} d u

\int \frac{u}{u^{2} - a u + b} d u = \int (\frac{2 u - a}{2 (u^{2} - a u + b)} + \frac{a}{2 (u^{2} - a u + b}) d u =

= \frac{1}{2} \int \frac{2 u - a}{u^{2} - a u + b} d u + \frac{a}{2} \int \frac{d u}{u^{2} - a u + b} =

[these are standard integrals]

= \frac{1}{2} \ln (u^{2} - a u + b) + \frac{a}{\sqrt{4 b - a^{2}}} \arctan \frac{2 u - a}{\sqrt{4 b - a^{2}}}

\int \frac{u}{u^{2} + a u + b} d u = \int (\frac{2 u + a}{2 (u^{2} + a u + b)} - \frac{a}{2 (u^{2} + a u + b)}) d u =

= \frac{1}{2} \int \frac{2 u + a}{u^{2} + a u + b} d u - \frac{a}{2} \int \frac{d u}{u^{2} + a u + b} =

[again standard integrals]

= \frac{1}{2} \ln (u^{2} + a u + b) - \frac{a}{\sqrt{4 b - a^{2}}} \arctan \frac{2 u + a}{\sqrt{4 b - a^{2}}}

= \frac{1}{2 a} (\ln (u^{2} - a u + b) + \ln (u^{2} + a u + b)) + \frac{1}{\sqrt{4 b - a^{2}}} (\arctan \frac{2 u - a}{\sqrt{4 b - a^{2}}} - \arctan \frac{2 u + a}{\sqrt{4 b - a^{2}}}) =

= \frac{1}{2 a} \ln (u^{4} + (2 b - a^{2}) u^{2} + b^{2}) + \frac{1}{\sqrt{4 b - a^{2}}} (\arctan \frac{2 u - a}{\sqrt{4 b - a^{2}}} - \arctan \frac{2 u + a}{\sqrt{4 b - a^{2}}}) =

= \frac{1}{2 a} \ln ∣ t^{2} + (2 b - a^{2}) t + b^{2} ∣ + \frac{1}{\sqrt{4 b - a^{2}}} (\arctan \frac{2 \sqrt{t} - a}{\sqrt{4 b - a^{2}}} - \arctan \frac{2 \sqrt{t} + a}{\sqrt{4 b - a^{2}}}) + C

sorry no time to finish

t = 1 + \tan θ

a = \sqrt{2 + \sqrt{8}}

b = \sqrt{2}

Commented by maxmathsup by imad last updated on 13/Sep/18

t h a n k y o u s i r .

t h a n k y o u s i r .

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