find the value of ∫_0 ^(π/4) ((tan(x)dx)/((√2)cos(x) +2sin^2 (x)))


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Question Number 40143 by maxmathsup by imad last updated on 16/Jul/18

$f i n d t h e v a l u e o f \int_{0}^{\frac{π}{4}} \frac{t a n (x) d x}{\sqrt{2} c o s (x) + 2 {s i n}^{2} (x)}$
Commented by math khazana by abdo last updated on 22/Jul/18
$I = ∫_0 ^(π/4) ((sinx)/(cosx((√2)cosx +2(1−cos^2 x)))dx changement cosx =t give I = ∫_1 ^(1/(√2)) ((√(1−t^2 ))/(t{(√2)t +2−2t^2 })) ((−dt)/(√(1−t^2 ))) = ∫_(1/(√2)) ^1 (dt/(t{−2t^2 +(√2)t +2})) let decompose F(t) = (1/(t{ −2t^2 +(√2) t+2})) Δ^ =2−4(−2)2=2 +16 =18 ⇒ t_1 =((−(√2) +3(√2))/(−2)) =(((√2) −3(√2))/2)=−(√2) t_2 =((−(√2) −3(√2))/(−2)) =(((√2)+3(√2))/2)=2(√2) F(t) = (1/(−2t(t−t_1 )(t−t_2 ))) =(a/t) +(b/(t−t_1 )) +(c/(t−t_2 )) a=lim_(t→0) tF(t)=(1/2) b =lim_(t→t_1 ) (t−t_1 )F(t) = ((−1)/(2t_1 (t_1 −t_2 ))) = ((−1)/(−2(√2)(−3(√2)))) = (1/(12)) c = lim_(t→t_2 ) (t−t_2 )F(t)= ((−1)/(2t_2 (t_2 −t_1 ))) =((−1)/(4(√2)(3(√2)))) =((−1)/(24)) ⇒ F(t) = (1/(2t)) +(1/(12(t−t_1 ))) −(1/(24(t−t_2 ))) ⇒ ∫_(1/(√2)) ^1 F(t)dt =(1/2) ∫_(1/(√2)) ^1 (dt/t) +(1/(12)) ∫_(1/(√2)) ^1 (dt/(t+(√2))) −(1/(24)) ∫_(1/(√2)) ^1 (dt/(t−2(√2))) =(1/2)[ln∣t∣]_(1/(√2)) ^1 +(1/(12))[ln∣t+(√2)∣]_(1/(√2)) ^1 −(1/(24))[ln∣t−2(√2)∣]_(1/(√2)) ^1 =((ln((√(2))))/2) +(1/(12)){ln(1+(√2))−ln((√2) +(1/(√2)))} −(1/(24)){ln(2(√2) −1)−ln(2(√2) −(1/(√2)))} .$
$I = \int_{0}^{\frac{π}{4}} \frac{s i n x}{c o s x (\sqrt{2} c o s x + 2 (1 - {c o s}^{2} x)} d x$ $c h a n g e m e n t c o s x = t g i v e$ $I = \int_{1}^{\frac{1}{\sqrt{2}}} \frac{\sqrt{1 - t^{2}}}{t {\sqrt{2} t + 2 - 2 t^{2}}} \frac{- d t}{\sqrt{1 - t^{2}}}$ $= \int_{\frac{1}{\sqrt{2}}}^{1} \frac{d t}{t {- 2 t^{2} + \sqrt{2} t + 2}} l e t d e c o m p o s e$ $F (t) = \frac{1}{t {- 2 t^{2} + \sqrt{2} t + 2}}$ $Δ^{} = 2 - 4 (- 2) 2 = 2 + 16 = 18 \Rightarrow$ $t_{1} = \frac{- \sqrt{2} + 3 \sqrt{2}}{- 2} = \frac{\sqrt{2} - 3 \sqrt{2}}{2} = - \sqrt{2}$ $t_{2} = \frac{- \sqrt{2} - 3 \sqrt{2}}{- 2} = \frac{\sqrt{2} + 3 \sqrt{2}}{2} = 2 \sqrt{2}$ $F (t) = \frac{1}{- 2 t (t - t_{1}) (t - t_{2})} = \frac{a}{t} + \frac{b}{t - t_{1}} + \frac{c}{t - t_{2}}$ $a = {l i m}_{t \to 0} t F (t) = \frac{1}{2}$ $b = {l i m}_{t \to t_{1}} (t - t_{1}) F (t) = \frac{- 1}{2 t_{1} (t_{1} - t_{2})}$ $= \frac{- 1}{- 2 \sqrt{2} (- 3 \sqrt{2})} = \frac{1}{12}$ $c = {l i m}_{t \to t_{2}} (t - t_{2}) F (t) = \frac{- 1}{2 t_{2} (t_{2} - t_{1})} = \frac{- 1}{4 \sqrt{2} (3 \sqrt{2})}$ $= \frac{- 1}{24} \Rightarrow$ $F (t) = \frac{1}{2 t} + \frac{1}{12 (t - t_{1})} - \frac{1}{24 (t - t_{2})} \Rightarrow$ $\int_{\frac{1}{\sqrt{2}}}^{1} F (t) d t = \frac{1}{2} \int_{\frac{1}{\sqrt{2}}}^{1} \frac{d t}{t} + \frac{1}{12} \int_{\frac{1}{\sqrt{2}}}^{1} \frac{d t}{t + \sqrt{2}} - \frac{1}{24} \int_{\frac{1}{\sqrt{2}}}^{1} \frac{d t}{t - 2 \sqrt{2}}$ $= \frac{1}{2} {[l n ∣ t ∣]}_{\frac{1}{\sqrt{2}}}^{1} + \frac{1}{12} {[l n ∣ t + \sqrt{2} ∣]}_{\frac{1}{\sqrt{2}}}^{1} - \frac{1}{24} {[l n ∣ t - 2 \sqrt{2} ∣]}_{\frac{1}{\sqrt{2}}}^{1}$ $= \frac{l n (\sqrt{2)}}{2} + \frac{1}{12} {l n (1 + \sqrt{2}) - l n (\sqrt{2} + \frac{1}{\sqrt{2}})}$ $- \frac{1}{24} {l n (2 \sqrt{2} - 1) - l n (2 \sqrt{2} - \frac{1}{\sqrt{2}})} .$
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