let I =∫_0 ^(π/4) e^(−2t) cos^4 t dt and J=∫_0 ^(π/4) e^(−2t) sin^4 tdt 1)calculate I+J and I−J 2) find the value of I and J.


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Question Number 66334 by mathmax by abdo last updated on 12/Aug/19

$l e t I = \int_{0}^{\frac{π}{4}} e^{- 2 t} {c o s}^{4} t d t a n d J = \int_{0}^{\frac{π}{4}} e^{- 2 t} {s i n}^{4} t d t$ $1) c a l c u l a t e I + J a n d I - J$ $2) f i n d t h e v a l u e o f I a n d J .$
Commented by mathmax by abdo last updated on 13/Aug/19
$1) I+J =∫_0 ^(π/4) (cos^4 t +sin^4 t)e^(−2t) dt =∫_0 ^(π/4) ((cos^2 t+sin^2 t)^2 −2sin^2 tcos^2 t)e^(−2t) dt =∫_0 ^(π/4) (1−(1/2)sin^2 (2t))e^(−2t) dt =∫_0 ^(π/4) e^(−2t) dt−(1/4)∫_0 ^(π/4) (1−cos(4t))e^(−2t) dt =(3/4) ∫_0 ^(π/4) e^(−2t) dt +(1/4) ∫_0 ^(π/4) cos(4t)e^(−2t) dt ∫_0 ^(π/4) e^(−2t) dt =[−(1/2)e^(−2t) ]_0 ^(π/4) =−(1/2)(e^(−(π/2)) −1) ∫_0 ^(π/4) cos(4t)e^(−2t) dt =Re(∫_0 ^(π/4) e^(−2t+i4t) dt) ∫_0 ^(π/4) e^((−2+4i)t) dt =[(1/(−2+4i))e^((−2+4i)t) ]_0 ^(π/4) =−(1/(2−4i)) { e^((−2+4i)(π/4)) −1} =−(((2+4i))/(20)){ e^(−(π/2)) (cos(π)+isin(π))−1} =((1+2i)/(10)){ 1+e^(−(π/2)) } ⇒∫_0 ^(π/4) cos(4t)e^(−2t) dt =(1/(10))e^(−(π/2)) ⇒ I +J =−(3/8)(e^(−(π/2)) −1) +(1/(40)) e^(−(π/2)) =((1/(40))−(3/8))e^(−(π/2)) +(3/8) =−((14)/(40))e^(−(π/2)) +(3/8) =−(7/(20))e^(−(π/2)) +(3/8)$
$1) I + J = \int_{0}^{\frac{π}{4}} ({c o s}^{4} t + {s i n}^{4} t) e^{- 2 t} d t = \int_{0}^{\frac{π}{4}} ({({c o s}^{2} t + {s i n}^{2} t)}^{2} - 2 {s i n}^{2} {t c o s}^{2} t) e^{- 2 t} d t$ $= \int_{0}^{\frac{π}{4}} (1 - \frac{1}{2} {s i n}^{2} (2 t)) e^{- 2 t} d t = \int_{0}^{\frac{π}{4}} e^{- 2 t} d t - \frac{1}{4} \int_{0}^{\frac{π}{4}} (1 - c o s (4 t)) e^{- 2 t} d t$ $= \frac{3}{4} \int_{0}^{\frac{π}{4}} e^{- 2 t} d t + \frac{1}{4} \int_{0}^{\frac{π}{4}} c o s (4 t) e^{- 2 t} d t$ $\int_{0}^{\frac{π}{4}} e^{- 2 t} d t = {[- \frac{1}{2} e^{- 2 t}]}_{0}^{\frac{π}{4}} = - \frac{1}{2} (e^{- \frac{π}{2}} - 1)$ $\int_{0}^{\frac{π}{4}} c o s (4 t) e^{- 2 t} d t = R e (\int_{0}^{\frac{π}{4}} e^{- 2 t + i 4 t} d t)$ $\int_{0}^{\frac{π}{4}} e^{(- 2 + 4 i) t} d t = {[\frac{1}{- 2 + 4 i} e^{(- 2 + 4 i) t}]}_{0}^{\frac{π}{4}} = - \frac{1}{2 - 4 i} {e^{(- 2 + 4 i) \frac{π}{4}} - 1}$ $= - \frac{(2 + 4 i)}{20} {e^{- \frac{π}{2}} (c o s (π) + i s i n (π)) - 1}$ $= \frac{1 + 2 i}{10} {1 + e^{- \frac{π}{2}}} \Rightarrow \int_{0}^{\frac{π}{4}} c o s (4 t) e^{- 2 t} d t = \frac{1}{10} e^{- \frac{π}{2}} \Rightarrow$ $I + J = - \frac{3}{8} (e^{- \frac{π}{2}} - 1) + \frac{1}{40} e^{- \frac{π}{2}} = (\frac{1}{40} - \frac{3}{8}) e^{- \frac{π}{2}} + \frac{3}{8}$ $= - \frac{14}{40} e^{- \frac{π}{2}} + \frac{3}{8} = - \frac{7}{20} e^{- \frac{π}{2}} + \frac{3}{8}$
Commented by mathmax by abdo last updated on 13/Aug/19
$we have I−J =∫_0 ^(π/4) (cos^4 t−sin^4 t)e^(−2t) dt =∫_0 ^(π/4) (cos^2 t−sin^2 t)e^(−2t) dt =∫_0 ^(π/4) cos(2t)e^(−2t) dt =Re(∫_0 ^(π/4) e^(−2t+i2t) dt) and ∫_0 ^(π/4) e^((−2+2i)t) dt =[(1/(−2+2i))e^((−2+2i)t) ]_0 ^(π/4) =−(1/(2−2i)){ e^((−2+2i)(π/4)) −1} =−((2+2i)/8){ e^(−(π/2)) i −1} =((1+i)/4)(1−i e^(−(π/2)) ) =((1−ie^(−(π/2)) +i+e^(−(π/2)) )/4) =((1+e^(−(π/2)) +i(1−e^(−(π/2)) ))/4) ⇒ ∫_0 ^(π/4) cos(2t)e^(−2t) dt =(1/4)(1+e^(−(π/2)) ) ⇒ I−J =(1/4) +(1/4)e^(−(π/2)) we have I+J =(3/8)−(7/(20))e^(−(π/2)) ⇒2I =(1/4)+(3/8) +((1/4)−(7/(20)))e^(−(π/2)) =(5/8) −(1/(10))e^(−(π/2)) also 2J =(3/8)−(1/4) +(−(7/(20))−(1/4))e^(−(π/2)) =(1/8) −(3/5)e^(−(π/2)) ⇒ I =(5/(16)) −(1/(20))e^(−(π/2)) and J =(1/(16)) −(3/(10))e^(−(π/2)) .$
$w e h a v e I - J = \int_{0}^{\frac{π}{4}} ({c o s}^{4} t - {s i n}^{4} t) e^{- 2 t} d t = \int_{0}^{\frac{π}{4}} ({c o s}^{2} t - {s i n}^{2} t) e^{- 2 t} d t$ $= \int_{0}^{\frac{π}{4}} c o s (2 t) e^{- 2 t} d t = R e (\int_{0}^{\frac{π}{4}} e^{- 2 t + i 2 t} d t) a n d$ $\int_{0}^{\frac{π}{4}} e^{(- 2 + 2 i) t} d t = {[\frac{1}{- 2 + 2 i} e^{(- 2 + 2 i) t}]}_{0}^{\frac{π}{4}} = - \frac{1}{2 - 2 i} {e^{(- 2 + 2 i) \frac{π}{4}} - 1}$ $= - \frac{2 + 2 i}{8} {e^{- \frac{π}{2}} i - 1} = \frac{1 + i}{4} (1 - i e^{- \frac{π}{2}})$ $= \frac{1 - {i e}^{- \frac{π}{2}} + i + e^{- \frac{π}{2}}}{4} = \frac{1 + e^{- \frac{π}{2}} + i (1 - e^{- \frac{π}{2}})}{4} \Rightarrow$ $\int_{0}^{\frac{π}{4}} c o s (2 t) e^{- 2 t} d t = \frac{1}{4} (1 + e^{- \frac{π}{2}}) \Rightarrow I - J = \frac{1}{4} + \frac{1}{4} e^{- \frac{π}{2}}$ $w e h a v e I + J = \frac{3}{8} - \frac{7}{20} e^{- \frac{π}{2}} \Rightarrow 2 I = \frac{1}{4} + \frac{3}{8} + (\frac{1}{4} - \frac{7}{20}) e^{- \frac{π}{2}}$ $= \frac{5}{8} - \frac{1}{10} e^{- \frac{π}{2}} a l s o 2 J = \frac{3}{8} - \frac{1}{4} + (- \frac{7}{20} - \frac{1}{4}) e^{- \frac{π}{2}}$ $= \frac{1}{8} - \frac{3}{5} e^{- \frac{π}{2}} \Rightarrow$ $I = \frac{5}{16} - \frac{1}{20} e^{- \frac{π}{2}} a n d J = \frac{1}{16} - \frac{3}{10} e^{- \frac{π}{2}} .$
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