1) calculate f(x)=∫_0 ^(π/4) ln(1+xtanθ)dθ 2) find the values of integrals ∫_0 ^(π/4) ln(1+tanθ) and ∫_0 ^(π/4) ln(1+2tanθ)dθ . 1) we have f^′ (x)=∫_0 ^(π/4) ((tanθ)/(1+xtanθ)) dθ =∫_0 ^(π/4) (((sinθ)/(cosθ))/(1+x((sinθ)/(cosθ))))dθ =∫_0 ^(π/4) ((sinθ)/(cosθ +xsinθ)) dθ =_(tan((θ/2))=t) ∫_0 ^((√2)−1) (((2t)/(1+t^2 ))/(((1−t^2 )/(1+t^2 )) +((2xt)/(1+t^2 )))) ((2dt)/(1+t^2 )) =∫_0 ^((√2)−1) ((4t)/((1+t^2 )(1−t^2 +2xt)))dt =−∫_0 ^((√2)−1) ((4t)/((t^2 +1)(t^2 −2xt −1)))dt let decompose F(t) = ((4t)/((t^2 +1)(t^2 −2xt −1))) roots of t^2 −2xt −1 Δ^′ =x^2 +1 ⇒t_1 =x+(√(x^2 +1)) and t_2 =x−(√(x^2 +1)) F(t)=(a/(t−t_1 )) +(b/(t−t_2 )) +((ct +d)/(t^2 +1)) a =lim_(t→t_1 ) (t−t_1 )F(t)=((4t_1 )/((t_1 ^2 +1)(t_1 −t_2 ))) =α b =lim_(t→t_2 ) (t−t_2 )F(t) =((4t_2 )/((t_2 ^2 +1)(t_2 −t_1 ))) =β ⇒F(t)=(α/(t−t_1 )) +(β/(t−t_2 )) +((ct +d)/(t^2 +1)) F(0) =0=−(α/t_1 ) −(β/t_2 ) +d ⇒d =(α/t_1 ) +(β/t_2 ) F(1)=(2/(−2x)) =−(1/x)=(α/(1−t_1 )) +(β/(1−t_2 )) +((c+d)/2) ⇒(1/x) =(α/(t_1 −1)) +(β/(t_2 −1)) −(c/2) −(d/2) ⇒(c/2) =(α/(t_1 −1)) +(β/(t_2 −1)) −(d/2) −(1/x) ⇒c =((2α)/(t_1 −1)) +((2β)/(t_2 −1)) −d−(2/x) ∫ F(t)dt =αln∣t−t_1 ∣ +βln∣t−t_2 ∣ +(c/2)ln(t^2 +1) +d arctan(t) ⇒ ∫_0 ^((√2)−1) F(t)dt =[αln∣t−t_1 ∣+βln∣t−t_2 ∣ +(c/2)ln(t^2 +1)]_0 ^((√2)−1) =αln∣(√2)−1−t_1 ∣ +βln∣(√2)−1−t_2 ∣ +(c/2)ln(4−2(√2)) =αln∣(√2)−1−x−(√(1+x^2 )))+βln∣(√2)−1−x+(√(1+x^2 ))) +((ln(4−2(√2)))/2)c =f^′ (x) ⇒ f(x)=∫ αln∣(√2)−1−x−(√(1+x^2 ))∣)dx+β∫ ln∣(√2)−1+(√(1+x^2 ))∣dx +((cx)/2)ln(4−2(√2)) +C ....be continued...


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Question Number 55267 by maxmathsup by imad last updated on 25/Feb/19

$1) c a l c u l a t e f (x) = \int_{0}^{\frac{π}{4}} l n (1 + x t a n θ) d θ$ $2) f i n d t h e v a l u e s o f i n t e g r a l s \int_{0}^{\frac{π}{4}} l n (1 + t a n θ) a n d \int_{0}^{\frac{π}{4}} l n (1 + 2 t a n θ) d θ .$ $1) w e h a v e f^{^{'}} (x) = \int_{0}^{\frac{π}{4}} \frac{t a n θ}{1 + x t a n θ} d θ = \int_{0}^{\frac{π}{4}} \frac{\frac{s i n θ}{c o s θ}}{1 + x \frac{s i n θ}{c o s θ}} d θ$ $= \int_{0}^{\frac{π}{4}} \frac{s i n θ}{c o s θ + x s i n θ} d θ =_{t a n (\frac{θ}{2}) = t} \int_{0}^{\sqrt{2} - 1} \frac{\frac{2 t}{1 + t^{2}}}{\frac{1 - t^{2}}{1 + t^{2}} + \frac{2 x t}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}$ $= \int_{0}^{\sqrt{2} - 1} \frac{4 t}{(1 + t^{2}) (1 - t^{2} + 2 x t)} d t = - \int_{0}^{\sqrt{2} - 1} \frac{4 t}{(t^{2} + 1) (t^{2} - 2 x t - 1)} d t l e t d e c o m p o s e$ $F (t) = \frac{4 t}{(t^{2} + 1) (t^{2} - 2 x t - 1)} r o o t s o f t^{2} - 2 x t - 1$ $Δ^{^{'}} = x^{2} + 1 \Rightarrow t_{1} = x + \sqrt{x^{2} + 1} a n d t_{2} = x - \sqrt{x^{2} + 1}$ $F (t) = \frac{a}{t - t_{1}} + \frac{b}{t - t_{2}} + \frac{c t + d}{t^{2} + 1}$ $a = {l i m}_{t \to t_{1}} (t - t_{1}) F (t) = \frac{4 t_{1}}{(t_{1}^{2} + 1) (t_{1} - t_{2})} = α$ $b = {l i m}_{t \to t_{2}} (t - t_{2}) F (t) = \frac{4 t_{2}}{(t_{2}^{2} + 1) (t_{2} - t_{1})} = β \Rightarrow F (t) = \frac{α}{t - t_{1}} + \frac{β}{t - t_{2}} + \frac{c t + d}{t^{2} + 1}$ $F (0) = 0 = - \frac{α}{t_{1}} - \frac{β}{t_{2}} + d \Rightarrow d = \frac{α}{t_{1}} + \frac{β}{t_{2}}$ $F (1) = \frac{2}{- 2 x} = - \frac{1}{x} = \frac{α}{1 - t_{1}} + \frac{β}{1 - t_{2}} + \frac{c + d}{2} \Rightarrow \frac{1}{x} = \frac{α}{t_{1} - 1} + \frac{β}{t_{2} - 1} - \frac{c}{2} - \frac{d}{2}$ $\Rightarrow \frac{c}{2} = \frac{α}{t_{1} - 1} + \frac{β}{t_{2} - 1} - \frac{d}{2} - \frac{1}{x} \Rightarrow c = \frac{2 α}{t_{1} - 1} + \frac{2 β}{t_{2} - 1} - d - \frac{2}{x}$ $\int F (t) d t = α l n ∣ t - t_{1} ∣ + β l n ∣ t - t_{2} ∣ + \frac{c}{2} l n (t^{2} + 1) + d a r c t a n (t) \Rightarrow$ $\int_{0}^{\sqrt{2} - 1} F (t) d t = {[α l n ∣ t - t_{1} ∣ + β l n ∣ t - t_{2} ∣ + \frac{c}{2} l n (t^{2} + 1)]}_{0}^{\sqrt{2} - 1}$ $= α l n ∣ \sqrt{2} - 1 - t_{1} ∣ + β l n ∣ \sqrt{2} - 1 - t_{2} ∣ + \frac{c}{2} l n (4 - 2 \sqrt{2})$ $= α l n ∣ \sqrt{2} - 1 - x - \sqrt{1 + x^{2}}) + β l n ∣ \sqrt{2} - 1 - x + \sqrt{1 + x^{2}}) + \frac{l n (4 - 2 \sqrt{2})}{2} c = f^{^{'}} (x) \Rightarrow$ $f (x) = \int α l n ∣ \sqrt{2} - 1 - x - \sqrt{1 + x^{2}} ∣) d x + β \int l n ∣ \sqrt{2} - 1 + \sqrt{1 + x^{2}} ∣ d x$ $+ \frac{c x}{2} l n (4 - 2 \sqrt{2}) + C . . . . b e c o n t i n u e d . . .$
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