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Question Number 43840 by peter frank last updated on 16/Sep/18

Commented by maxmathsup by imad last updated on 16/Sep/18
$let I = ∫_0 ^(π/2) ((sin^4 x)/(cos^4 x +sin^4 x))dx and J = ∫_0 ^(π/2) ((cos^4 x)/(cos^4 x +sin^4 x))dx we have I +J = ∫_0 ^(π/2) dx =(π/2) and J −I =∫_0 ^(π/2) ((cos^4 x −sin^4 x)/(cos^4 x +sin^4 x))dx = ∫_0 ^(π/2) ((cos^2 x −sin^2 x)/((cos^2 x +sin^2 x)^2 −2cos^2 x sin^2 x))dx =∫_0 ^(π/2) ((cos(2x))/(1−(1/2)sin^2 (2x)))dx = ∫_0 ^(π/2) ((2cos(2x))/(2−(1−cos^2 (2x))))dx = ∫_0 ^(π/2) ((2cos(2x))/(1+cos^2 (2x)))dx =_(2x=t) ∫_0 ^π ((2cost)/(1+cos^2 t)) dt = _(tan((t/2))=u) ∫_0 ^∞ ((2 ((1−u^2 )/(1+u^2 )))/(1+(((1−u^2 )/(1+u^2 )))^2 )) ((2du)/(1+u^2 )) = 4 ∫_0 ^∞ ((1−u^2 )/((1+u^2 )^2 { (((1+u^2 )^2 +(1−u^2 )^2 )/((1+u^2 )^2 ))}))du = 4 ∫_0 ^∞ ((1−u^2 )/(u^4 +2u^2 +1 +u^4 −2u^2 +1))du =4 ∫_0 ^∞ ((1−u^2 )/(2(u^4 +1)))du = 2 ∫_0 ^∞ ((1−u^2 )/(1+u^4 )) du =2 ∫_0 ^∞ (du/(1+u^4 )) −2 ∫_0 ^∞ (u^2 /(1+u^4 )) du but ∫_0 ^∞ (du/(1+u^4 )) =(π/(2(√2))) changement u=α^(1/4) give ∫_0 ^∞ (α^(1/2) /(1+α)) (1/4) α^((1/4)−1) dα = (1/4) ∫_0 ^∞ (α^((3/4)−1) /(1+α)) dα =(1/4) (π/(sin(((3π)/4)))) =(π/(4 ((√2)/2))) =(π/(2(√2))) ⇒ J−I = (π/(√2)) −(π/(√2)) =0 ⇒I =J ⇒ 2I =(π/2) ⇒ I =(π/4) and J =(π/4) .$
$l e t I = \int_{0}^{\frac{π}{2}} \frac{{s i n}^{4} x}{{c o s}^{4} x + {s i n}^{4} x} d x a n d J = \int_{0}^{\frac{π}{2}} \frac{{c o s}^{4} x}{{c o s}^{4} x + {s i n}^{4} x} d x w e h a v e$ $I + J = \int_{0}^{\frac{π}{2}} d x = \frac{π}{2} a n d J - I = \int_{0}^{\frac{π}{2}} \frac{{c o s}^{4} x - {s i n}^{4} x}{{c o s}^{4} x + {s i n}^{4} x} d x$ $= \int_{0}^{\frac{π}{2}} \frac{{c o s}^{2} x - {s i n}^{2} x}{{({c o s}^{2} x + {s i n}^{2} x)}^{2} - 2 {c o s}^{2} x {s i n}^{2} x} d x = \int_{0}^{\frac{π}{2}} \frac{c o s (2 x)}{1 - \frac{1}{2} {s i n}^{2} (2 x)} d x$ $= \int_{0}^{\frac{π}{2}} \frac{2 c o s (2 x)}{2 - (1 - {c o s}^{2} (2 x))} d x = \int_{0}^{\frac{π}{2}} \frac{2 c o s (2 x)}{1 + {c o s}^{2} (2 x)} d x$ $=_{2 x = t} \int_{0}^{π} \frac{2 c o s t}{1 + {c o s}^{2} t} d t =_{t a n (\frac{t}{2}) = u} \int_{0}^{\infty} \frac{2 \frac{1 - u^{2}}{1 + u^{2}}}{1 + {(\frac{1 - u^{2}}{1 + u^{2}})}^{2}} \frac{2 d u}{1 + u^{2}}$ $= 4 \int_{0}^{\infty} \frac{1 - u^{2}}{{(1 + u^{2})}^{2} {\frac{{(1 + u^{2})}^{2} + {(1 - u^{2})}^{2}}{{(1 + u^{2})}^{2}}}} d u$ $= 4 \int_{0}^{\infty} \frac{1 - u^{2}}{u^{4} + 2 u^{2} + 1 + u^{4} - 2 u^{2} + 1} d u = 4 \int_{0}^{\infty} \frac{1 - u^{2}}{2 (u^{4} + 1)} d u$ $= 2 \int_{0}^{\infty} \frac{1 - u^{2}}{1 + u^{4}} d u = 2 \int_{0}^{\infty} \frac{d u}{1 + u^{4}} - 2 \int_{0}^{\infty} \frac{u^{2}}{1 + u^{4}} d u b u t$ $\int_{0}^{\infty} \frac{d u}{1 + u^{4}} = \frac{π}{2 \sqrt{2}} c h a n g e m e n t u = α^{\frac{1}{4}} g i v e$ $\int_{0}^{\infty} \frac{α^{\frac{1}{2}}}{1 + α} \frac{1}{4} α^{\frac{1}{4} - 1} d α = \frac{1}{4} \int_{0}^{\infty} \frac{α^{\frac{3}{4} - 1}}{1 + α} d α = \frac{1}{4} \frac{π}{s i n (\frac{3 π}{4})} = \frac{π}{4 \frac{\sqrt{2}}{2}} = \frac{π}{2 \sqrt{2}} \Rightarrow$ $J - I = \frac{π}{\sqrt{2}} - \frac{π}{\sqrt{2}} = 0 \Rightarrow I = J \Rightarrow 2 I = \frac{π}{2} \Rightarrow I = \frac{π}{4} a n d J = \frac{π}{4} .$
	Answered by Joel578 last updated on 16/Sep/18

	$\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ $I = \int_{0}^{\frac{π}{2}} \frac{\sin^{4} x}{\cos^{4} x + \sin^{4} x} d x . . . (i)$ $using formula above gives us :$ $I = \int_{0}^{\frac{π}{2}} \frac{\sin^{4} (\frac{π}{2} - x)}{\cos^{4} (\frac{π}{2} - x) + \sin^{4} (\frac{π}{2} - x)} d x$ $= \int_{0}^{\frac{π}{2}} \frac{\cos^{4} x}{\sin^{4} x + \cos^{4} x} d x . . . (i i)$ $(i) + (i i)$ $2 I = \int_{0}^{\frac{π}{2}} \frac{\sin^{4} x}{\cos^{4} x + \sin^{4} x} d x + \int_{0}^{\frac{π}{2}} \frac{\cos^{4} x}{\sin^{4} x + \cos^{4} x} d x$ $= \int_{0}^{\frac{π}{2}} 1 d x$ $= \frac{π}{2}$ $I = \frac{π}{4}$
	Commented by Joel578 last updated on 16/Sep/18

	$Thanks to Mr . Tanmay who taught me$ $this method$
	Commented by peter frank last updated on 16/Sep/18

	$t h a n k s f o r y o u r t i m e$
	Commented by tanmay.chaudhury50@gmail.com last updated on 16/Sep/18

	$t h a n k y o u . . . y o u r e t a i n e d w h a t y o u l e a r n e d . . .$
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