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


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 44308 by abdo.msup.com last updated on 26/Sep/18

$l e t I = \int_{0}^{\infty} {c o s}^{4} t e^{- 2 t} d t a n d J = \int_{0}^{\infty} {s i n}^{4} t e^{- 2 t} d t$ $1) c a l c u l a t e I + J a n d I - J$ $2) f i n d t h e v a l u e s o f I a n d J .$
Commented by tanmay.chaudhury50@gmail.com last updated on 27/Sep/18

Commented by maxmathsup by imad last updated on 28/Sep/18
$1) we have I +J = ∫_0 ^∞ (cos^4 t +sin^4 t)e^(−2t) dt =∫_0 ^∞ { (cos^2 t +sin^2 t)^2 −2cos^2 t sin^2 t}e^(−2t) dt = ∫_0 ^∞ {1−(1/2)sin^2 (2t)}e^(−2t) dt = ∫_0 ^∞ e^(−2t) dt −(1/2) ∫_0 ^∞ ( ((1−cos(2t))/2))e^(−2t) dt =[−(1/2)e^(−2t) ]_0 ^(+∞) −(1/4) ∫_0 ^∞ e^(−2t) dt +(1/4) ∫_0 ^∞ e^(−2t) cos(2t)dt =(1/2) +(1/8)[ e^(−2t) ]_0 ^(+∞) +(1/4)∫_0 ^∞ e^(−2t) cos(2t)dt=(3/8) +(1/4) Re(∫_0 ^∞ e^(−2t+i2t) dt) but ∫_0 ^∞ e^((−2+2i)t) dt =[ (1/(−2+2i)) e^((−2+2i)t) ]_0 ^(+∞) =((−1)/(−2+2i)) =(1/(2−2i)) = ((2+2i)/8) =((1+i)/4) ⇒ I +J =(3/8) +(1/(16)) =(7/(16)) also we have I−J =∫_0 ^∞ (cos^4 t −sin^4 t)e^(−2t) dt =∫_0 ^∞ (cos^2 t −sin^2 t)e^(−2t) dt =∫_0 ^∞ e^(−2t) cos(2t)dt =Re( ∫_0 ^∞ e^((−2+2i)t) dt)=(1/4) 2) we have I +J =(7/(16)) and I−J =(1/4) ⇒ 2I =(7/(16)) +(1/4) =((11)/(16)) ⇒I =((11)/(32)) and J =(7/(16)) −((11)/(32)) =((14−11)/(32)) =(3/(32)) .$
$1) w e h a v e I + J = \int_{0}^{\infty} ({c o s}^{4} t + {s i n}^{4} t) e^{- 2 t} d t$ $= \int_{0}^{\infty} {{({c o s}^{2} t + {s i n}^{2} t)}^{2} - 2 {c o s}^{2} t {s i n}^{2} t} e^{- 2 t} d t$ $= \int_{0}^{\infty} {1 - \frac{1}{2} {s i n}^{2} (2 t)} e^{- 2 t} d t = \int_{0}^{\infty} e^{- 2 t} d t - \frac{1}{2} \int_{0}^{\infty} (\frac{1 - c o s (2 t)}{2}) e^{- 2 t} d t$ $= {[- \frac{1}{2} e^{- 2 t}]}_{0}^{+ \infty} - \frac{1}{4} \int_{0}^{\infty} e^{- 2 t} d t + \frac{1}{4} \int_{0}^{\infty} e^{- 2 t} c o s (2 t) d t$ $= \frac{1}{2} + \frac{1}{8} {[e^{- 2 t}]}_{0}^{+ \infty} + \frac{1}{4} \int_{0}^{\infty} e^{- 2 t} c o s (2 t) d t = \frac{3}{8} + \frac{1}{4} R e (\int_{0}^{\infty} e^{- 2 t + i 2 t} d t)$ $b u t \int_{0}^{\infty} e^{(- 2 + 2 i) t} d t = {[\frac{1}{- 2 + 2 i} e^{(- 2 + 2 i) t}]}_{0}^{+ \infty} = \frac{- 1}{- 2 + 2 i} = \frac{1}{2 - 2 i}$ $= \frac{2 + 2 i}{8} = \frac{1 + i}{4} \Rightarrow I + J = \frac{3}{8} + \frac{1}{16} = \frac{7}{16} a l s o w e h a v e$ $I - J = \int_{0}^{\infty} ({c o s}^{4} t - {s i n}^{4} t) e^{- 2 t} d t = \int_{0}^{\infty} ({c o s}^{2} t - {s i n}^{2} t) e^{- 2 t} d t$ $= \int_{0}^{\infty} e^{- 2 t} c o s (2 t) d t = R e (\int_{0}^{\infty} e^{(- 2 + 2 i) t} d t) = \frac{1}{4}$ $2) w e h a v e I + J = \frac{7}{16} a n d I - J = \frac{1}{4} \Rightarrow 2 I = \frac{7}{16} + \frac{1}{4} = \frac{11}{16} \Rightarrow I = \frac{11}{32}$ $a n d J = \frac{7}{16} - \frac{11}{32} = \frac{14 - 11}{32} = \frac{3}{32} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 27/Sep/18
	$I+J=∫_0 ^∞ e^(−2t) (cos^4 t+sin^4 t)dt =∫_0 ^∞ e^(−2t) {(cos^2 t+sin^2 t)^2 −2sin^2 tcos^2 t}dt =∫_0 ^∞ e^(−2t) {1−((sin^2 2t)/2)}dt =∫_0 ^∞ e^(−2t) {1−((1−cos4t)/4)} =∫_0 ^∞ e^(−2t) (((4−1+cos4t)/4)} =(3/4)∫_0 ^∞ e^(−2t) dt+(1/4)∫_0 ^∞ e^(−2t) cos4t dt formula ∫e^(ax) cosbx dx=((e^(ax) (acosbx+bsinbx))/(a^2 +b^2 )) =(3/4)∣(e^(−2t) /(−2))∣_0 ^∞ +(1/4)∣((e^(−2t) (−2cos4t+4sin4t))/((−2)^2 +(4)^2 ))∣_0 ^∞ I−J=∫_0 ^∞ e^(−2t) (cos^4 t−sin^4 t)dt =∫_0 ^∞ e^(−2t) cos2t dt =∣((e^(−2t) (−2cos2t+2sin2t)/((−2)^2 +(4^2 )))∣_0 ^∞ pls check upper limit of the interval...$
	$I + J = \int_{0}^{\infty} e^{- 2 t} ({c o s}^{4} t + {s i n}^{4} t) d t$ $= \int_{0}^{\infty} e^{- 2 t} {{({c o s}^{2} t + {s i n}^{2} t)}^{2} - 2 {s i n}^{2} {t c o s}^{2} t} d t$ $= \int_{0}^{\infty} e^{- 2 t} {1 - \frac{{s i n}^{2} 2 t}{2}} d t$ $= \int_{0}^{\infty} e^{- 2 t} {1 - \frac{1 - c o s 4 t}{4}}$ $= \int_{0}^{\infty} e^{- 2 t} (\frac{4 - 1 + c o s 4 t}{4}}$ $= \frac{3}{4} \int_{0}^{\infty} e^{- 2 t} d t + \frac{1}{4} \int_{0}^{\infty} e^{- 2 t} c o s 4 t d t$ $f o r m u l a \int e^{a x} c o s b x d x = \frac{e^{a x} (a c o s b x + b s i n b x)}{a^{2} + b^{2}}$ $= \frac{3}{4} ∣ \frac{e^{- 2 t}}{- 2} ∣_{0}^{\infty} + \frac{1}{4} ∣ \frac{e^{- 2 t} (- 2 c o s 4 t + 4 s i n 4 t)}{{(- 2)}^{2} + {(4)}^{2}} ∣_{0}^{\infty}$ $I - J = \int_{0}^{\infty} e^{- 2 t} ({c o s}^{4} t - {s i n}^{4} t) d t$ $= \int_{0}^{\infty} e^{- 2 t} c o s 2 t d t$ $=∣ \frac{e^{- 2 t} (- 2 c o s 2 t + 2 s i n 2 t}{{(- 2)}^{2} + (4^{2})} ∣_{0}^{\infty}$ $p l s c h e c k u p p e r l i m i t o f t h e i n t e r v a l . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum