let-I-0-pi-4-e-2t-cos-4-t-dt-and-J-0-pi-4-e-2t-sin-4-tdt-1-calculate-I-J-and-I-J-2-find-the-value-of-I-and-J-

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 = \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

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}} .