let f(x) =∫_0 ^(π/4) (dt/(x+tant)) with x real 1) find aexplicit form of f(x) 2)find also g(x) =∫_0 ^(π/4) (dt/((x+tant)^2 )) 3)give f^((n)) (x)at form of integral 4)calculate ∫_0 ^(π/4) (dt/(2+tant)) and ∫


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Question Number 66060 by mathmax by abdo last updated on 08/Aug/19

$l e t f (x) = \int_{0}^{\frac{π}{4}} \frac{d t}{x + t a n t} w i t h x r e a l$ $1) f i n d a e x p l i c i t f o r m o f f (x)$ $2) f i n d a l s o g (x) = \int_{0}^{\frac{π}{4}} \frac{d t}{{(x + t a n t)}^{2}}$ $3) g i v e f^{(n)} (x) a t f o r m o f i n t e g r a l$ $4) c a l c u l a t e \int_{0}^{\frac{π}{4}} \frac{d t}{2 + t a n t} a n d \int_{0}^{\frac{π}{4}} \frac{d t}{{(2 + t a n t)}^{2}}$
Commented by mathmax by abdo last updated on 10/Aug/19

$1) w e h a v e f (x) = \int_{0}^{\frac{π}{4}} \frac{d t}{x + t a n t} c h a n g e m e n t t a n (\frac{t}{2}) = u g i v e$ $f (x) = \int_{0}^{\sqrt{2} - 1} \frac{1}{x + \frac{2 u}{1 - u^{2}}} \frac{2 d u}{1 + u^{2}} = 2 \int_{0}^{\sqrt{2} - 1} \frac{1 - u^{2}}{(1 + u^{2}) (x - {x u}^{2} + 2 u)} d u$ $= 2 \int_{0}^{\sqrt{2} - 1} \frac{u^{2} - 1}{(u^{2} + 1) ({x u}^{2} - 2 u - x)} d u l e t d e c o m p o s e$ $F (u) = \frac{u^{2} - 1}{(u^{2} + 1) ({x u}^{2} - 2 u - x)}$ ${x u}^{2} - 2 u - x = 0 \to Δ^{^{'}} = 1 + x^{2} \Rightarrow u_{1} = \frac{1 + \sqrt{1 + x^{2}}}{x} a n d u_{2} = \frac{1 - \sqrt{1 + x^{2}}}{x}$ $(w e s u p p o s e x \neq 0) \Rightarrow F (u) = \frac{u^{2} - 1}{x (u - u_{1}) (x - u_{2}) (u^{2} + 1)}$ $= \frac{a}{u - u_{1}} + \frac{b}{u - u_{2}} + \frac{c u + d}{u^{2} + 1}$ $a = {l i m}_{u \to u_{1}} (u - u_{1}) F (u) = \frac{u_{1}^{2} - 1}{x (2 \frac{\sqrt{1 + x^{2}}}{x}) (u_{1}^{2} + 1)} = \frac{u_{1}^{2} - 1}{2 \sqrt{1 + x^{2}} (u_{1}^{2} + 1)}$ $b = {l i m}_{u \to u_{2}} (u - u_{2}) F (u) = \frac{u_{2}^{2} - 1}{x (- 2 \frac{\sqrt{1 + x^{2}}}{x}) (u_{2}^{2} + 1)} = \frac{u_{2}^{2} - 1}{- 2 \sqrt{1 + x^{2}} (u_{2}^{2} + 1)}$ ${l i m}_{u \to + \infty} u F (u) = 0 = a + b + c \Rightarrow c = - (a + b)$ $F (0) = \frac{1}{x} = - \frac{a}{u_{1}} - \frac{b}{u_{2}} + d \Rightarrow d = \frac{1}{x} + \frac{a}{u_{1}} + \frac{b}{u_{2}} \Rightarrow$ $\int_{0}^{\sqrt{2} - 1} F (u) d u = {[a l n ∣ u - u_{1} ∣ + b l n ∣ u - u_{2} ∣]}_{1}^{\sqrt{2} - 1} + {[\frac{c}{2} l n (u^{2} + 1) + d a r c t a n (u)]}_{0}^{\sqrt{2} - 1}$ $= a l n ∣ \frac{\sqrt{2} - 1 - u_{1}}{1 - u_{1}} ∣ + b l n ∣ \frac{\sqrt{2} - 1 - u_{2}}{1 - u_{2}} ∣ + \frac{c}{2} l n (3 - 2 \sqrt{2} + 1) + d a r c t a n (\sqrt{2} - 1)}$ $\Rightarrow f (x) = 2 a l n ∣ \frac{\sqrt{2} - 1 - u_{1}}{1 - u_{1}} ∣ + 2 b l n ∣ \frac{\sqrt{2} - 1 - u_{2}}{1 - u_{2}} ∣ + c l n (4 - 2 \sqrt{2}) + \frac{d π}{4}$ $a = \frac{{(\frac{1 + \sqrt{1 + x^{2}}}{x})}^{2} - 1}{2 \sqrt{1 + x^{2}} ({(\frac{1 + \sqrt{1 + x^{2}}}{x})}^{2} + 1)} = \frac{{(1 + \sqrt{1 + x^{2}})}^{2} - x^{2}}{2 \sqrt{1 + x^{2}} ({(1 + \sqrt{1 + x^{2}})}^{2} + x^{2})}$ $= \frac{1 + 2 \sqrt{1 + x^{2}} + 1 + x^{2} - x^{2}}{2 \sqrt{1 + x^{2}} (1 + 2 \sqrt{1 + x^{2}} + 1 + x^{2} + x^{2})} = \frac{(1 + \sqrt{1 + x^{2}})}{\sqrt{1 + x^{2}} (2 + 2 \sqrt{1 + x^{2}} + 2 x^{2})}$ $= \frac{1 + \sqrt{1 + x^{2}}}{2 \sqrt{1 + x^{2}} (x^{2} + 2 \sqrt{1 + x^{2}} + 1)} r e s t t o s i m p l i f y o t h e r c o e f f i c i e 7 t s . . .$
Commented by mathmax by abdo last updated on 10/Aug/19

$2) w e h a v e f^{^{'}} (x) = - \int_{0}^{\frac{π}{4}} \frac{d t}{{(x + t a n t)}^{2}} = - g (x) \Rightarrow g (x) = - f^{^{'}} (x)$ $r e s t t o c a l c u l a t e f^{^{'}} (x)$ $3) w e h a v e f (x) = \int_{0}^{\frac{π}{4}} \frac{d t}{(x + t a n t)} \Rightarrow f^{(n)} (x) = \int_{0}^{\frac{π}{4}} \frac{{(- 1)}^{n} n!}{{(x + t a n t)}^{n + 1}} d t$
Commented by mathmax by abdo last updated on 10/Aug/19
$4) changement tan((t/2))=u give ∫_0 ^(π/4) (dt/((2+tant))) =∫_0 ^((√2)−1) (1/(2+((2u)/(1−u^2 ))))((2du)/(1+u^2 )) =2 ∫_0 ^((√2)−1) ((1−u^2 )/((1+u^2 )(2−2u^2 +2u))) =2 ∫_0 ^((√2)−1) ((u^2 −1)/((u^2 +1)(2u^2 −2u−2))) =∫_0 ^((√2)−1) ((u^2 −1)/((u^2 −u−1)(u^2 +1)))du let decompose F(u) =((u^2 −1)/((u^2 −u−1)(u^2 +1))) u^2 −u−1=0→Δ =1+4 =5 ⇒u_1 =((1+(√5))/2) and u_2 =((1−(√5))/2) F(u)=((u^2 −1)/((u−u_1 )(u−u_2 )(u^2 +1))) =(a/(u−u_1 )) +(b/(u−u_2 )) +((cu +d)/(u^2 +1)) a =((u_1 ^2 −1)/((√5)(u_1 ^2 +1))) =(((((1+(√5))^2 )/4)−1)/((√5)( (((1+(√5))^2 )/4)+1))) =((6+2(√5)−4)/((√5)(6+2(√5)+4))) =((2+2(√5))/((√5)(10+2(√5)))) =((1+(√5))/((√5)(5+(√5)))) =((1+(√5))/(5(√5)+5)) =(1/5) b =((u_2 ^2 −1)/(−(√5)(u_2 ^2 +1))) =(((((1−(√5))^2 )/4)−1)/((−(√5))((((1−(√5))^2 )/4)+1))) =((6−2(√5)−4)/((−(√5))(6−2(√5)+4))) =((2−2(√5))/((−(√5))(10−2(√5)))) =((1−(√5))/((−(√5))(5−(√5)))) =(((√5)−1)/(5(√5)−5)) =(1/5) lim_(u→+∞) uF(u)=0 =a+b+c ⇒c =−(2/5) F(0) =1 =−(a/u_1 )−(b/u_2 ) +d ⇒d =1+(1/(5u_1 )) +(1/(5u_2 )) =1+(1/5){((u_1 +u_2 )/(u_1 u_2 ))} =1+(1/5) (1/(−1)) =1−(1/5) =(4/5) ⇒ F(u)= (1/(5(u−u_1 ))) +(1/(5(u−u_2 ))) +((((−2)/5)u +(4/5))/(u^2 +1)) ⇒ ∫ F(u)du =(1/5)ln∣u−u_1 ∣+(1/5)ln∣u−u_2 ∣−(1/5)ln(u^2 +1)+(4/5) arctanu +c ⇒ ∫_0 ^((√2)−1) F(u)du= [(1/5)ln∣u−u_1 ∣+(1/5)ln∣u−u_2 ∣−(1/5)ln(u^2 +1)+(4/5)arctan(u)]_0 ^((√2)−1) ...rest to finish the calculus...$
$4) c h a n g e m e n t t a n (\frac{t}{2}) = u g i v e$ $\int_{0}^{\frac{π}{4}} \frac{d t}{(2 + t a n t)} = \int_{0}^{\sqrt{2} - 1} \frac{1}{2 + \frac{2 u}{1 - u^{2}}} \frac{2 d u}{1 + u^{2}} = 2 \int_{0}^{\sqrt{2} - 1} \frac{1 - u^{2}}{(1 + u^{2}) (2 - 2 u^{2} + 2 u)}$ $= 2 \int_{0}^{\sqrt{2} - 1} \frac{u^{2} - 1}{(u^{2} + 1) (2 u^{2} - 2 u - 2)} = \int_{0}^{\sqrt{2} - 1} \frac{u^{2} - 1}{(u^{2} - u - 1) (u^{2} + 1)} d u$ $l e t d e c o m p o s e F (u) = \frac{u^{2} - 1}{(u^{2} - u - 1) (u^{2} + 1)}$ $u^{2} - u - 1 = 0 \to Δ = 1 + 4 = 5 \Rightarrow u_{1} = \frac{1 + \sqrt{5}}{2} a n d u_{2} = \frac{1 - \sqrt{5}}{2}$ $F (u) = \frac{u^{2} - 1}{(u - u_{1}) (u - u_{2}) (u^{2} + 1)} = \frac{a}{u - u_{1}} + \frac{b}{u - u_{2}} + \frac{c u + d}{u^{2} + 1}$ $a = \frac{u_{1}^{2} - 1}{\sqrt{5} (u_{1}^{2} + 1)} = \frac{\frac{{(1 + \sqrt{5})}^{2}}{4} - 1}{\sqrt{5} (\frac{{(1 + \sqrt{5})}^{2}}{4} + 1)} = \frac{6 + 2 \sqrt{5} - 4}{\sqrt{5} (6 + 2 \sqrt{5} + 4)} = \frac{2 + 2 \sqrt{5}}{\sqrt{5} (10 + 2 \sqrt{5})}$ $= \frac{1 + \sqrt{5}}{\sqrt{5} (5 + \sqrt{5})} = \frac{1 + \sqrt{5}}{5 \sqrt{5} + 5} = \frac{1}{5}$ $b = \frac{u_{2}^{2} - 1}{- \sqrt{5} (u_{2}^{2} + 1)} = \frac{\frac{{(1 - \sqrt{5})}^{2}}{4} - 1}{(- \sqrt{5}) (\frac{{(1 - \sqrt{5})}^{2}}{4} + 1)} = \frac{6 - 2 \sqrt{5} - 4}{(- \sqrt{5}) (6 - 2 \sqrt{5} + 4)}$ $= \frac{2 - 2 \sqrt{5}}{(- \sqrt{5}) (10 - 2 \sqrt{5})} = \frac{1 - \sqrt{5}}{(- \sqrt{5}) (5 - \sqrt{5})} = \frac{\sqrt{5} - 1}{5 \sqrt{5} - 5} = \frac{1}{5}$ ${l i m}_{u \to + \infty} u F (u) = 0 = a + b + c \Rightarrow c = - \frac{2}{5}$ $F (0) = 1 = - \frac{a}{u_{1}} - \frac{b}{u_{2}} + d \Rightarrow d = 1 + \frac{1}{5 u_{1}} + \frac{1}{5 u_{2}}$ $= 1 + \frac{1}{5} {\frac{u_{1} + u_{2}}{u_{1} u_{2}}} = 1 + \frac{1}{5} \frac{1}{- 1} = 1 - \frac{1}{5} = \frac{4}{5} \Rightarrow$ $F (u) = \frac{1}{5 (u - u_{1})} + \frac{1}{5 (u - u_{2})} + \frac{\frac{- 2}{5} u + \frac{4}{5}}{u^{2} + 1} \Rightarrow$ $\int F (u) d u = \frac{1}{5} l n ∣ u - u_{1} ∣ + \frac{1}{5} l n ∣ u - u_{2} ∣ - \frac{1}{5} l n (u^{2} + 1) + \frac{4}{5} a r c t a n u + c$ $\Rightarrow \int_{0}^{\sqrt{2} - 1} F (u) d u =$ ${[\frac{1}{5} l n ∣ u - u_{1} ∣ + \frac{1}{5} l n ∣ u - u_{2} ∣ - \frac{1}{5} l n (u^{2} + 1) + \frac{4}{5} a r c t a n (u)]}_{0}^{\sqrt{2} - 1}$ $. . . r e s t t o f i n i s h t h e c a l c u l u s . . .$
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