calculate I = ∫_0 ^(π/4) ((cosx)/(cos^3 x +sin^3 x))dx


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Question Number 41703 by abdo.msup.com last updated on 11/Aug/18

$c a l c u l a t e I = \int_{0}^{\frac{π}{4}} \frac{c o s x}{{c o s}^{3} x + {s i n}^{3} x} d x$
Commented by math khazana by abdo last updated on 12/Aug/18
$I = ∫_0 ^(π/4) (((cosx)/(cos^3 x))/((cos^3 x +sin^3 x)/(cos^3 x))) dx =∫_0 ^(π/4) (1/(cos^2 x(1+tan^3 x)))dx =∫_0 ^(π/4) ((1+tan^2 x)/(1+tan^3 x)) dx =_(tanx=t) ∫_0 ^1 ((1+t^2 )/(1+t^3 )) (dt/(1+t^2 )) = ∫_0 ^1 (dt/(1+t^3 )) let decompose F(t)= (1/(t^3 +1)) =(1/((t+1)(t^2 −t +1))) F(t)=(a/(t+1)) +((bt +c)/(t^2 −t +1)) a =lim_(t→−1) (t+1)F(t) =(1/3) lim_(t→+∞) tF(t)=0=a +b ⇒b=−(1/3) ⇒ F(t) = (1/(3(t+1))) +((−(1/3)t +c)/(t^2 −t +1)) F(0) =1 =(1/3) +c ⇒c=(2/3) ⇒ F(t) = (1/(3(t+1))) −(1/3) ((t−2)/(t^2 −t +1)) ⇒ ∫_0 ^1 F(t)dt =(1/3)[ln∣t+1∣]_0 ^1 −(1/3) ∫_0 ^1 ((t−2)/(t^2 −t +1))dt =((ln(2))/3) −(1/6) ∫_0 ^1 ((2t−1−3)/(t^2 −t +1)) dt =((ln(2))/3) −[(1/6)ln∣t^2 −t+1∣ ]_0 ^1 +(1/2) ∫_0 ^1 (dt/(t^2 −t +1)) =((ln(2))/3) +(1/2) ∫_0 ^1 (dt/((t−(1/2))^2 +(3/4))) =_(t−(1/2)=((√3)/2)u) ((ln(2))/3) +(1/2) (4/3) ∫_(−(1/(√3))) ^(1/(√3)) (1/(1+u^2 )) ((√3)/2) du =((ln(2))/3) + ((√3)/3) [ arctanu]_(−(1/(√3))) ^(1/(√3)) =((ln(2))/3) + (1/(√3)){ 2 arctan((1/(√3)))} =((ln(2))/3) +(2/(√3)) (π/6) ⇒ I = ((ln(2))/3) +(π/(3(√3))) .$
$I = \int_{0}^{\frac{π}{4}} \frac{\frac{c o s x}{{c o s}^{3} x}}{\frac{{c o s}^{3} x + {s i n}^{3} x}{{c o s}^{3} x}} d x = \int_{0}^{\frac{π}{4}} \frac{1}{{c o s}^{2} x (1 + {t a n}^{3} x)} d x$ $= \int_{0}^{\frac{π}{4}} \frac{1 + {t a n}^{2} x}{1 + {t a n}^{3} x} d x =_{t a n x = t} \int_{0}^{1} \frac{1 + t^{2}}{1 + t^{3}} \frac{d t}{1 + t^{2}}$ $= \int_{0}^{1} \frac{d t}{1 + t^{3}} l e t d e c o m p o s e$ $F (t) = \frac{1}{t^{3} + 1} = \frac{1}{(t + 1) (t^{2} - t + 1)}$ $F (t) = \frac{a}{t + 1} + \frac{b t + c}{t^{2} - t + 1}$ $a = {l i m}_{t \to - 1} (t + 1) F (t) = \frac{1}{3}$ ${l i m}_{t \to + \infty} t F (t) = 0 = a + b \Rightarrow b = - \frac{1}{3} \Rightarrow$ $F (t) = \frac{1}{3 (t + 1)} + \frac{- \frac{1}{3} t + c}{t^{2} - t + 1}$ $F (0) = 1 = \frac{1}{3} + c \Rightarrow c = \frac{2}{3} \Rightarrow$ $F (t) = \frac{1}{3 (t + 1)} - \frac{1}{3} \frac{t - 2}{t^{2} - t + 1} \Rightarrow$ $\int_{0}^{1} F (t) d t = \frac{1}{3} {[l n ∣ t + 1 ∣]}_{0}^{1} - \frac{1}{3} \int_{0}^{1} \frac{t - 2}{t^{2} - t + 1} d t$ $= \frac{l n (2)}{3} - \frac{1}{6} \int_{0}^{1} \frac{2 t - 1 - 3}{t^{2} - t + 1} d t$ $= \frac{l n (2)}{3} - {[\frac{1}{6} l n ∣ t^{2} - t + 1 ∣]}_{0}^{1} + \frac{1}{2} \int_{0}^{1} \frac{d t}{t^{2} - t + 1}$ $= \frac{l n (2)}{3} + \frac{1}{2} \int_{0}^{1} \frac{d t}{{(t - \frac{1}{2})}^{2} + \frac{3}{4}}$ $=_{t - \frac{1}{2} = \frac{\sqrt{3}}{2} u} \frac{l n (2)}{3} + \frac{1}{2} \frac{4}{3} \int_{- \frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} \frac{1}{1 + u^{2}} \frac{\sqrt{3}}{2} d u$ $= \frac{l n (2)}{3} + \frac{\sqrt{3}}{3} {[a r c t a n u]}_{- \frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}}$ $= \frac{l n (2)}{3} + \frac{1}{\sqrt{3}} {2 a r c t a n (\frac{1}{\sqrt{3}})}$ $= \frac{l n (2)}{3} + \frac{2}{\sqrt{3}} \frac{π}{6} \Rightarrow I = \frac{l n (2)}{3} + \frac{π}{3 \sqrt{3}} .$
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