let-I-0-pi-8-e-2t-cos-4-t-and-J-0-pi-8-e-2t-sin-4-dt-find-the-values-of-I-andJ-

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

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}} \dots 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}} \dots . b e c o n t i n u e d \dots

Commented by maxmathsup by imad last updated on 06/Sep/18

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}}