let f(x) =∫_(π/3) ^(π/2) (dθ/(1+xtanθ)) with x real 1) find a explicit form for f(x) 2) determine also g(x) =∫_(π/3) ^(π/2) ((tanθ)/((1+xtanθ)^2 )) dθ 3) let U_n (x) =f^((n)) (x) give U_n (x) at form of integral. 4) calculate ∫_(π/3) ^(π/2) (dθ/(1+2tanθ)) and ∫


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Question Number 58774 by maxmathsup by imad last updated on 29/Apr/19

$l e t f (x) = \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{d θ}{1 + x t a n θ} w i t h x r e a l$ $1) f i n d a e x p l i c i t f o r m f o r f (x)$ $2) d e t e r m i n e a l s o g (x) = \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{t a n θ}{{(1 + x t a n θ)}^{2}} d θ$ $3) l e t U_{n} (x) = f^{(n)} (x) g i v e U_{n} (x) a t f o r m o f i n t e g r a l .$ $4) c a l c u l a t e \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{d θ}{1 + 2 t a n θ} a n d \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{t a n θ d θ}{{(1 + 2 t a n θ)}^{2}}$
Commented by maxmathsup by imad last updated on 01/May/19
$1) we have f(x) =∫_(π/3) ^(π/2) (dθ/(1+x ((2tan((θ/2)))/(1−tan^2 ((θ/2)))))) =_(tan((θ/2))=t) ∫_(1/(√3)) ^1 (1/(1+((2xt)/(1−t^2 )))) ((2dt)/(1+t^2 )) = ∫_(1/(√3)) ^1 ((2(1−t^2 )dt)/((1−t^2 +2xt)(1+t^2 ))) =∫_(1/(√3)) ^1 ((2(t^2 −1))/((t^2 −2xt−1)(t^2 +1))) dt roots of t^2 −2xt −1 →Δ^′ =x^2 +1⇒ t_1 =x+(√(1+x^2 )) and t_2 =x−(√(1+x^2 )) let decompose F(t) =((2(t^2 −1))/((t^2 −2xt −1)(t^2 +1))) ⇒F(t) =(a/(t−t_1 )) +(b/(t−t_2 )) +((ct +d)/(t^2 +1)) =((2(t^2 −1))/((t−t_1 )(t−t_2 )(t^2 +1))) a =lim_(t→t_1 ) (t−t_1 )F(t) =((2(t_1 ^2 −1))/((t_1 −t_2 )(t_1 ^2 +1))) =((2(t_1 ^2 −1))/(2(√(1+x^2 ))(t_1 ^2 +1))) b =lim_(t→t_2 ) (t−t_2 )F(t) =((2(t_2 ^2 −1))/(−2(√(1+x^2 ))(t_2 ^2 +1))) lim_(t→+∞) tF(t) =0 =a+b +c ⇒c =−a−b ⇒F(t) =(a/(t−t_1 )) +(b/(t−t_2 )) +(((−a−b)t +d)/(t^2 +1)) F(0) =2 =−(a/t_1 ) −(b/t_2 ) +d ⇒d =(a/t_1 ) +(b/t_2 ) +2 ⇒ ∫_(1/(√3)) ^1 F(t)dt =a∫_(1/(√3)) ^1 (dt/(t−t_1 )) +b ∫_(1/(√3)) ^1 (dt/(t−t_2 )) −((a+b)/2) ∫_(1/(√3)) ^1 ((2dt)/(t^2 +1)) +d ∫_(1/(√3)) ^1 (dt/(t^2 +1)) =[aln∣t−t_1 ∣ +bln∣t−t_2 ∣]_(1/(√3)) ^1 −((a+b)/2)[ln(t^2 +1)]_(1/(√3)) ^1 + d [arctan(t)]_(1/(√3)) ^1 =aln∣1−t_1 ∣+bln∣1−t_2 ∣−aln∣(1/(√3)) −t_1 ∣−bln∣(1/(√3))−t_2 ∣−((a+b)/2){ln(2)−ln((4/3))} +d {(π/4) −(π/6)}=aln∣((1−t_1 )/((1/(√3))−t_1 ))∣ +bln∣((1−t_2 )/((1/(√3))−t_2 ))∣ −((a+b)/2)(ln(3)−ln(2)) +((πd)/(12)) a =((2( (x+(√(1+x^2 )))^2 −1))/(2(√(1+x^2 ))( (x+(√(1+x^2 )))^2 +1))) =(({ x^2 +2x(√(1+x^2 )) + x^2 })/((√(1+x^2 ))(x^2 +2x(√(1+x^2 )) +x^2 +2))) =(({x^2 +x(√(1+x^2 ))})/((√(1+x^2 )){x^2 +x(√(1+x^2 +1})))) b =((2( (x−(√(1+x^2 )))^2 −1))/(−2(√(1+x^2 ))( (x−(√(1+x^2 )))^2 +1))) =− ((x^2 −2x(√(1+x^2 )) +x^2 )/((√(1+x^2 ))(x^2 −2x(√(1+x^2 )) +x^2 +2))) =−(((x^2 −x(√(1+x^2 ))))/((√(1+x^2 ))( x^2 −x(√(1+x^2 +1))+1))) ....be continued....$
$1) w e h a v e f (x) = \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{d θ}{1 + x \frac{2 t a n (\frac{θ}{2})}{1 - {t a n}^{2} (\frac{θ}{2})}} =_{t a n (\frac{θ}{2}) = t} \int_{\frac{1}{\sqrt{3}}}^{1} \frac{1}{1 + \frac{2 x t}{1 - t^{2}}} \frac{2 d t}{1 + t^{2}}$ $= \int_{\frac{1}{\sqrt{3}}}^{1} \frac{2 (1 - t^{2}) d t}{(1 - t^{2} + 2 x t) (1 + t^{2})} = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{2 (t^{2} - 1)}{(t^{2} - 2 x t - 1) (t^{2} + 1)} d t r o o t s o f$ $t^{2} - 2 x t - 1 \to Δ^{^{'}} = x^{2} + 1 \Rightarrow t_{1} = x + \sqrt{1 + x^{2}} a n d t_{2} = x - \sqrt{1 + x^{2}} l e t d e c o m p o s e$ $F (t) = \frac{2 (t^{2} - 1)}{(t^{2} - 2 x t - 1) (t^{2} + 1)} \Rightarrow F (t) = \frac{a}{t - t_{1}} + \frac{b}{t - t_{2}} + \frac{c t + d}{t^{2} + 1} = \frac{2 (t^{2} - 1)}{(t - t_{1}) (t - t_{2}) (t^{2} + 1)}$ $a = {l i m}_{t \to t_{1}} (t - t_{1}) F (t) = \frac{2 (t_{1}^{2} - 1)}{(t_{1} - t_{2}) (t_{1}^{2} + 1)} = \frac{2 (t_{1}^{2} - 1)}{2 \sqrt{1 + x^{2}} (t_{1}^{2} + 1)}$ $b = {l i m}_{t \to t_{2}} (t - t_{2}) F (t) = \frac{2 (t_{2}^{2} - 1)}{- 2 \sqrt{1 + x^{2}} (t_{2}^{2} + 1)}$ ${l i m}_{t \to + \infty} t F (t) = 0 = a + b + c \Rightarrow c = - a - b \Rightarrow F (t) = \frac{a}{t - t_{1}} + \frac{b}{t - t_{2}} + \frac{(- a - b) t + d}{t^{2} + 1}$ $F (0) = 2 = - \frac{a}{t_{1}} - \frac{b}{t_{2}} + d \Rightarrow d = \frac{a}{t_{1}} + \frac{b}{t_{2}} + 2 \Rightarrow$ $\int_{\frac{1}{\sqrt{3}}}^{1} F (t) d t = a \int_{\frac{1}{\sqrt{3}}}^{1} \frac{d t}{t - t_{1}} + b \int_{\frac{1}{\sqrt{3}}}^{1} \frac{d t}{t - t_{2}} - \frac{a + b}{2} \int_{\frac{1}{\sqrt{3}}}^{1} \frac{2 d t}{t^{2} + 1} + d \int_{\frac{1}{\sqrt{3}}}^{1} \frac{d t}{t^{2} + 1}$ $= {[a l n ∣ t - t_{1} ∣ + b l n ∣ t - t_{2} ∣]}_{\frac{1}{\sqrt{3}}}^{1} - \frac{a + b}{2} {[l n (t^{2} + 1)]}_{\frac{1}{\sqrt{3}}}^{1} + d {[a r c t a n (t)]}_{\frac{1}{\sqrt{3}}}^{1}$ $= a l n ∣ 1 - t_{1} ∣ + b l n ∣ 1 - t_{2} ∣ - a l n ∣ \frac{1}{\sqrt{3}} - t_{1} ∣ - b l n ∣ \frac{1}{\sqrt{3}} - t_{2} ∣ - \frac{a + b}{2} {l n (2) - l n (\frac{4}{3})}$ $+ d {\frac{π}{4} - \frac{π}{6}} = a l n ∣ \frac{1 - t_{1}}{\frac{1}{\sqrt{3}} - t_{1}} ∣ + b l n ∣ \frac{1 - t_{2}}{\frac{1}{\sqrt{3}} - t_{2}} ∣ - \frac{a + b}{2} (l n (3) - l n (2)) + \frac{π d}{12}$ $a = \frac{2 ({(x + \sqrt{1 + x^{2}})}^{2} - 1)}{2 \sqrt{1 + x^{2}} ({(x + \sqrt{1 + x^{2}})}^{2} + 1)} = \frac{{x^{2} + 2 x \sqrt{1 + x^{2}} + x^{2}}}{\sqrt{1 + x^{2}} (x^{2} + 2 x \sqrt{1 + x^{2}} + x^{2} + 2)}$ $= \frac{{x^{2} + x \sqrt{1 + x^{2}}}}{\sqrt{1 + x^{2}} {x^{2} + x \sqrt{1 + x^{2} + 1}}}$ $b = \frac{2 ({(x - \sqrt{1 + x^{2}})}^{2} - 1)}{- 2 \sqrt{1 + x^{2}} ({(x - \sqrt{1 + x^{2}})}^{2} + 1)} = - \frac{x^{2} - 2 x \sqrt{1 + x^{2}} + x^{2}}{\sqrt{1 + x^{2}} (x^{2} - 2 x \sqrt{1 + x^{2}} + x^{2} + 2)}$ $= - \frac{(x^{2} - x \sqrt{1 + x^{2}})}{\sqrt{1 + x^{2}} (x^{2} - x \sqrt{1 + x^{2} + 1} + 1)} . . . . b e c o n t i n u e d . . . .$
Commented by maxmathsup by imad last updated on 01/May/19

$2) w e h a v e f^{^{'}} (x) = - \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{t a n θ d θ}{{(1 + x t a n θ)}^{2}} = - g (x) \Rightarrow g (x) = - f^{^{'}} (x) r e s t t o c a l c u l a t e$ $f^{^{'}} (x)$ $3) w e h a v e f^{(1)} (x) = (- 1) \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{t a n θ}{{(1 + x t a n θ)}^{2}} d θ \Rightarrow f^{(2)} (x) = {(- 1)}^{2} \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{2 (1 + x t a n θ)}{{(1 + x t a n θ)}^{4}} {t a n}^{2} θ d θ$ $= 2 . {(- 1)}^{2} \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{{t a n}^{2} θ}{{(1 + x t a n θ)}^{3}} d θ l e t s u p p o s e f^{(n)} (x) = {(- 1)}^{n} n! \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{{t a n}^{n} θ}{{(1 + x t a n θ)}^{n + 1}} d θ$ $\Rightarrow f^{(n + 1)} (x) = {(- 1)}^{n + 1} n! \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{(n + 1) t a n θ {(1 + x t a n θ)}^{n} {t a n}^{n} θ}{{(1 + x t a n θ)}^{2 n + 2}} d θ$ $= {(- 1)}^{n + 1} (n + 1)! \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{{t a n}^{n + 1} θ}{{(1 + x t a n θ)}^{n + 2}} d θ s o t h e r e s u l t i s p r o v e d .$
Commented by maxmathsup by imad last updated on 01/May/19

$4) l e t A = \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{d θ}{1 + 2 t a n θ} \Rightarrow A = \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{d θ}{1 + 2 \frac{2 t a n (\frac{θ}{2})}{1 - {t a n}^{2} (\frac{θ}{2})}}$ $A =_{t a n (\frac{θ}{2}) = t} \int_{\frac{1}{\sqrt{3}}}^{1} \frac{1}{1 + \frac{4 t}{1 - t^{2}}} \frac{2 d t}{1 + t^{2}} = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{2 (1 - t^{2})}{(1 - t^{2} + 4 t) (1 + t^{2})} d t$ $= \int_{\frac{1}{\sqrt{3}}}^{1} \frac{2 (t^{2} - 1)}{(t^{2} - 4 t - 1) (t^{2} + 1)} d t l e t d e c o m p o s e F (t) = \frac{2 (t^{2} - 1)}{(t^{2} - 4 t - 1) (t^{2} + 1)}$ $r o o t f t^{2} - 4 t - 1 \to Δ^{^{'}} = 4 + 1 = 5 \Rightarrow t_{1} = 2 + \sqrt{5} a n d t_{2} = 2 - \sqrt{5} \Rightarrow$ $F (t) = \frac{a}{t - t_{1}} + \frac{b}{t - t_{2}} + \frac{c t + d}{t^{2} + 1} = \frac{2 (t^{2} - 1)}{(t - t_{1}) (t - t_{2}) (t^{2} + 1)}$ $a = {l i m}_{t \to t_{1}} (t - t_{1}) F (t) = \frac{2 (t_{1}^{2} - 1)}{2 \sqrt{5} (t_{1}^{2} + 1)} = \frac{{(2 + \sqrt{5})}^{2} - 1}{\sqrt{5} ({(2 + \sqrt{5})}^{2} + 1)}$ $= \frac{4 + 4 \sqrt{5} + 5 - 1}{\sqrt{5} (4 + 4 \sqrt{5} + 5 + 1)} = \frac{8 + 4 \sqrt{5}}{\sqrt{5} (10 + 4 \sqrt{5})} = \frac{4 + 2 \sqrt{5}}{\sqrt{5} (5 + 2 \sqrt{5})}$ $b = {l i m}_{t \to t_{2}} (t - t_{2}) F (t) = \frac{2 (t_{2}^{2} - 1)}{- 2 \sqrt{5} (t_{2}^{2} + 1)} = - \frac{{(2 - \sqrt{5})}^{2} - 1}{\sqrt{5} ({(2 - \sqrt{5})}^{2} + 1)}$ $= - \frac{4 - 4 \sqrt{5} + 5 - 1}{\sqrt{5} (4 - 4 \sqrt{5} + 5 + 1)} = - \frac{8 - 4 \sqrt{5}}{\sqrt{5} (10 - 4 \sqrt{5})} = - \frac{4 - 2 \sqrt{5}}{\sqrt{5} (5 - 2 \sqrt{5})} = \frac{- 4 + 2 \sqrt{5}}{\sqrt{5} (5 - 2 \sqrt{5})}$ ${l i m}_{t \to + \infty} t F (t) = 0 = a + b + c \Rightarrow c = - (a + b)$ $F (0) = - \frac{a}{t_{1}} - \frac{b}{t_{2}} + d = 2 \Rightarrow d = 2 + \frac{a}{t_{1}} + \frac{b}{t_{2}} \Rightarrow$ $\int_{\frac{1}{\sqrt{3}}}^{1} F (t) d t = a \int_{\frac{1}{\sqrt{3}}}^{1} \frac{d t}{t - t_{1}} + b \int_{\frac{1}{\sqrt{3}}}^{1} \frac{d t}{t - t_{2}} + \frac{c}{2} \int_{\frac{1}{\sqrt{3}}}^{1} \frac{2 t}{t^{2} + 1} d t + d \int_{\frac{1}{\sqrt{3}}}^{1} \frac{d t}{1 + t^{2}}$ $= {[a l n ∣ t - t_{1} ∣ + b l n ∣ t - t_{2} ∣ + \frac{c}{2} l n (t^{2} + 1) + d a r c t a n (t)]}_{\frac{1}{\sqrt{3}}}^{1}$ $= a l n ∣ 1 - t_{1} ∣ + b l n ∣ 1 - t_{2} ∣ + \frac{c}{2} l n (2) + \frac{d π}{4} - a l n ∣ \frac{1}{\sqrt{3}} - t_{1} ∣ - b l n ∣ \frac{1}{\sqrt{3}} - t_{2} ∣$ $- \frac{c}{2} l n (\frac{4}{3}) - \frac{d π}{6} . . . .$
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