let I = ∫_0 ^(π/8) e^(−2t) cos^4 t and J =∫_0 ^(π/8) e^(−2t) sin^4 dt find the values of I andJ .


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Question Number 42799 by maxmathsup by imad last updated on 02/Sep/18

$l e t I = \int_{0}^{\frac{π}{8}} e^{- 2 t} {c o s}^{4} t a n d J = \int_{0}^{\frac{π}{8}} e^{- 2 t} {s i n}^{4} d t$ $f i n d t h e v a l u e s o f I a n d J .$
Commented by maxmathsup by imad last updated on 06/Sep/18
$we have I +J = ∫_0 ^(π/8) e^(−2t) { cos^4 t +sin^4 t}dt = ∫_0 ^(π/8) e^(−2t) { (cos^2 t +sin^2 t)^2 −2cos^2 t sin^2 t}dt =∫_0 ^(π/8) e^(−2t) { 1−(1/2) sin^2 2t} dt = ∫_0 ^(π/8) e^(−2t) dt −(1/2) ∫_0 ^(π/8) e^(−2t) sin^2 (2t) dt =−(1/2)[ e^(−2t) ]_0 ^(π/8) −(1/4) ∫_0 ^(π/8) e^(−2t) (1−cos(2t))dt =(1/2){1−e^(−(π/4)) } +(1/8)[ e^(−2t) ]_0 ^(π/8) +(1/4) ∫_0 ^(π/8) e^(−2t) cos(2t)dt =(1/2){1−e^(−(π/4)) } +(1/8){ e^(−(π/4)) −1} +(1/4) Re( ∫_0 ^(π/8) e^(−2t +2it) dt) but ∫_0 ^(π/8) e^((−2+2i)t) dt =[ (1/(−2+2i)) e^((−2+2i)t) ]_0 ^(π/8) =(1/(−2 +2i)) { e^((−2+2i)(π/8)) −1} = −(1/2) (1/(1−i)){ e^(−(π/4)) e^(i(π/4)) −1}} =−(1/2) (((1+i))/2)){ e^(−(π/4)) ( (1/(√2)) +(i/(√2))) −1} =((−1)/4)(1+i) { (e^(−(π/4)) /(√2)) + i (e^(−(π/4)) /(√2)) −1} =−(1/(4(√2))){ e^(−(π/4)) +i e^(−(π/4)) −(√2)}(1+i) =−(1/(4(√2))){ e^(−(π/4)) +i e^(−(π/4)) +i e^(−(π/4)) −e^(−(π/4)) −(√2)−(√2)i} = −(1/(4(√2))){ 2i e^(−(π/4)) −(√2)i −(√2)} ⇒Re( ∫_0 ^(π/2) ...dx)= (1/4) ⇒ I +J =(1/2) −(1/8) −(3/8) e^(−(π/4)) +(1/(16)) = (7/(16)) −(3/8) e^(−(π/4)) .... be continued...$
$w e h a v e I + J = \int_{0}^{\frac{π}{8}} e^{- 2 t} {{c o s}^{4} t + {s i n}^{4} t} d t$ $= \int_{0}^{\frac{π}{8}} e^{- 2 t} {{({c o s}^{2} t + {s i n}^{2} t)}^{2} - 2 {c o s}^{2} t {s i n}^{2} t} d t$ $= \int_{0}^{\frac{π}{8}} e^{- 2 t} {1 - \frac{1}{2} {s i n}^{2} 2 t} d t = \int_{0}^{\frac{π}{8}} e^{- 2 t} d t - \frac{1}{2} \int_{0}^{\frac{π}{8}} e^{- 2 t} {s i n}^{2} (2 t) d t$ $= - \frac{1}{2} {[e^{- 2 t}]}_{0}^{\frac{π}{8}} - \frac{1}{4} \int_{0}^{\frac{π}{8}} e^{- 2 t} (1 - c o s (2 t)) d t$ $= \frac{1}{2} {1 - e^{- \frac{π}{4}}} + \frac{1}{8} {[e^{- 2 t}]}_{0}^{\frac{π}{8}} + \frac{1}{4} \int_{0}^{\frac{π}{8}} e^{- 2 t} c o s (2 t) d t$ $= \frac{1}{2} {1 - e^{- \frac{π}{4}}} + \frac{1}{8} {e^{- \frac{π}{4}} - 1} + \frac{1}{4} R e (\int_{0}^{\frac{π}{8}} e^{- 2 t + 2 i t} d t) b u t$ $\int_{0}^{\frac{π}{8}} e^{(- 2 + 2 i) t} d t = {[\frac{1}{- 2 + 2 i} e^{(- 2 + 2 i) t}]}_{0}^{\frac{π}{8}} = \frac{1}{- 2 + 2 i} {e^{(- 2 + 2 i) \frac{π}{8}} - 1}$ $= - \frac{1}{2} \frac{1}{1 - i} {e^{- \frac{π}{4}} e^{i \frac{π}{4}} - 1}} = - \frac{1}{2} (\frac{1 + i)}{2}) {e^{- \frac{π}{4}} (\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}) - 1}$ $= \frac{- 1}{4} (1 + i) {\frac{e^{- \frac{π}{4}}}{\sqrt{2}} + i \frac{e^{- \frac{π}{4}}}{\sqrt{2}} - 1}$ $= - \frac{1}{4 \sqrt{2}} {e^{- \frac{π}{4}} + i e^{- \frac{π}{4}} - \sqrt{2}} (1 + i)$ $= - \frac{1}{4 \sqrt{2}} {e^{- \frac{π}{4}} + i e^{- \frac{π}{4}} + i e^{- \frac{π}{4}} - e^{- \frac{π}{4}} - \sqrt{2} - \sqrt{2} i}$ $= - \frac{1}{4 \sqrt{2}} {2 i e^{- \frac{π}{4}} - \sqrt{2} i - \sqrt{2}} \Rightarrow R e (\int_{0}^{\frac{π}{2}} . . . d x) = \frac{1}{4} \Rightarrow$ $I + J = \frac{1}{2} - \frac{1}{8} - \frac{3}{8} e^{- \frac{π}{4}} + \frac{1}{16} = \frac{7}{16} - \frac{3}{8} e^{- \frac{π}{4}} . . . . b e c o n t i n u e d . . .$
Commented by maxmathsup by imad last updated on 06/Sep/18
$we have I −J = ∫_0 ^(π/8) e^(−2t) {cos^4 t −sin^4 t}dt =∫_0 ^(π/8) e^(−2t) {cos^2 t −sin^2 t}dt = ∫_0 ^(π/8) e^(−2t) cos(2t)dt =Re( ∫_0 ^(π/8) e^(−2t +2it) dt) =Re( ∫_0 ^(π/8) e^((−2+2i)t) dt)=(1/4) so I+J =(7/(16)) −(3/8)e^(−(π/4)) and I−J=(1/4) ⇒ 2I =((11)/(16)) −(3/8) e^(−(π/4)) and 2J =(3/(16)) −(3/8) e^(−(π/4)) ⇒ I =((11)/(32)) −(3/(16)) e^(−(π/4)) and J =(3/(32)) −(3/(16)) e^(−(π/4))$
$w e h a v e I - J = \int_{0}^{\frac{π}{8}} e^{- 2 t} {{c o s}^{4} t - {s i n}^{4} t} d t$ $= \int_{0}^{\frac{π}{8}} e^{- 2 t} {{c o s}^{2} t - {s i n}^{2} t} d t = \int_{0}^{\frac{π}{8}} e^{- 2 t} c o s (2 t) d t = R e (\int_{0}^{\frac{π}{8}} e^{- 2 t + 2 i t} d t)$ $= R e (\int_{0}^{\frac{π}{8}} e^{(- 2 + 2 i) t} d t) = \frac{1}{4} s o I + J = \frac{7}{16} - \frac{3}{8} e^{- \frac{π}{4}} a n d I - J = \frac{1}{4} \Rightarrow$ $2 I = \frac{11}{16} - \frac{3}{8} e^{- \frac{π}{4}} a n d 2 J = \frac{3}{16} - \frac{3}{8} e^{- \frac{π}{4}} \Rightarrow$ $I = \frac{11}{32} - \frac{3}{16} e^{- \frac{π}{4}} a n d J = \frac{3}{32} - \frac{3}{16} e^{- \frac{π}{4}}$
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