calculate-A-0-pi-4-cos-8-xdx-and-B-0-pi-4-sin-8-xdx-2-calculate-A-B-and-A-B-3-calculate-A-2-B-2-

Question Number 41677 by math khazana by abdo last updated on 11/Aug/18

c a l c u l a t e A = \int_{0}^{\frac{π}{4}} {c o s}^{8} x d x a n d

B = \int_{0}^{\frac{π}{4}} {s i n}^{8} x d x

2) c a l c u l a t e A + B a n d A - B

3) c a l c u l a t e A^{2} - B^{2}

Commented by math khazana by abdo last updated on 12/Aug/18

l e t A_{n} = \int_{0}^{\frac{π}{4}} {c o s}^{2 n} x d x w e h a v e A = A_{4}

A_{n + 1} = \int_{0}^{\frac{π}{4}} {c o s}^{2 n} (1 - {s i n}^{2} x) d x

= A_{n} - \int_{0}^{\frac{π}{4}} {s i n}^{2} x {c o s}^{2 n} x d x b y p a r t s

\int_{0}^{\frac{π}{4}} s i n x (- s i n x) {c o s}^{2 n} x d x

= {[\frac{1}{2 n + 1} {s i n x c o s}^{2 n + 1} x]}_{0}^{\frac{π}{4}} - \int_{0}^{\frac{π}{4}} \frac{1}{2 n + 1} {c o s}^{2 n + 2} x d x

= \frac{1}{2 n + 1} {(\frac{1}{\sqrt{2}})}^{2 n + 2} - \frac{1}{2 n + 1} A_{n + 1} \Rightarrow

(1 + \frac{1}{2 n + 1}) A_{n + 1} = A_{n} + \frac{1}{(2 n + 1) 2^{n + 1}} \Rightarrow

\frac{2 n + 2}{2 n + 1} A_{n + 1} = A_{n} + \frac{1}{(2 n + 1) 2^{n + 1}} \Rightarrow

A_{n + 1} = \frac{2 n + 1}{2 n + 2} A_{n} + \frac{1}{(2 n + 2) 2^{n + 1}} \Rightarrow

A_{4} = \frac{7}{8} A_{3} + \frac{1}{8 . 2^{4}}

A_{3} = \frac{5}{6} A_{2} + \frac{1}{6 . 2^{3}}

A_{2} = \frac{3}{4} A_{1} + \frac{1}{4 . 2^{2}}

A_{1} = \frac{1}{2} A_{0} + \frac{1}{4} \Rightarrow

A_{4} = \frac{7}{8} {\frac{5}{6} A_{2} + \frac{1}{6 . 2^{3}}} + \frac{1}{8 . 2^{4}}

= \frac{35}{48} A_{2} + \frac{7}{48 . 2^{3}} + \frac{1}{8 . 2^{4}}

= \frac{35}{48} {\frac{3}{4} A_{1} + \frac{1}{4 . 2^{2}}} + \frac{7}{48 . 2^{3}} + \frac{1}{8 . 2^{4}}

= \frac{105}{192} A_{1} + \frac{35}{192 . 2^{2}} + \frac{7}{48 . 2^{3}} + \frac{1}{8 . 2^{4}}

= \frac{105}{192} {\frac{1}{2} A_{0} + \frac{1}{4}} + \frac{35}{192 . 2^{2}} + \frac{7}{48 . 2^{3}} + \frac{1}{8 . 2^{4}}

= \frac{105}{192} {\frac{π}{8} + \frac{1}{4}} + \frac{35}{192 . 2^{2}} + \frac{7}{48 . 2^{3}} + \frac{1}{8 . 2^{4}} = A

Commented by math khazana by abdo last updated on 12/Aug/18

l e t B_{n} = \int_{0}^{\frac{π}{4}} {s i n}^{2 n} x d x w e h a v e B = B_{4}

B_{n + 1} = \int_{0}^{\frac{π}{4}} {s i n}^{2 n} (1 - {c o s}^{2} x) d x

= \int_{0}^{\frac{π}{4}} {s i n}^{2 n} x d x - \int_{0}^{\frac{π}{4}} c o s x (c o s x) {s i n}^{2 n} x d x b y

p a r t s

\int_{0}^{\frac{π}{4}} c o s x (c o s x) {s i n}^{2 n} x d x = {[\frac{1}{2 n + 1} {c o s x s i n}^{2 n + 1} x]}_{0}^{\frac{π}{4}}

- \int_{0}^{\frac{π}{4}} (- s i n x) \frac{1}{2 n + 1} {s i n}^{2 n + 1} x d x

= \frac{1}{(2 n + 1)} {(\frac{1}{\sqrt{2}})}^{2 n + 2} + \frac{1}{2 n + 1} \int_{0}^{\frac{π}{4}} {s i n}^{2 n + 2} x d x

= \frac{1}{(2 n + 1) 2^{n + 1}} + \frac{1}{2 n + 1} B_{n + 1} \Rightarrow

B_{n + 1} = B_{n} - \frac{1}{(2 n + 1) 2^{n + 1}} - \frac{1}{2 n + 1} B_{n + 1} \Rightarrow

(1 + \frac{1}{2 n + 1}) B_{n + 1} = B_{n} - \frac{1}{(2 n + 1) 2^{n + 1}} \Rightarrow

\frac{2 n + 2}{2 n + 1} B_{n + 1} = B_{n} - \frac{1}{(2 n + 1) 2^{n + 1}} \Rightarrow

B_{n + 1} = \frac{2 n + 1}{2 n + 2} B_{n} - \frac{1}{(2 n + 2) 2^{n + 1}} \dots b e c o n t i n u e d \dots

Answered by ajfour last updated on 11/Aug/18

A = \int_{0}^{π / 4} \cos^{8} x d x

= \frac{1}{16} \int_{0}^{π / 4} {(1 + \cos 2 x)}^{4} d x

= \frac{1}{16} \int_{0}^{π / 4} (1 + 4 \cos 2 x + 6 \cos^{2} 2 x

+ 4 \cos^{3} 2 x + \cos^{4} 2 x) d x

= \frac{π}{64} + \frac{4 \sin 2 x}{32} ∣_{0}^{π / 4} + \frac{3}{16} \int_{0}^{π / 4} (1 + \cos 4 x) d x

+ \frac{1}{16} \int_{0}^{π / 4} (\cos 6 x + 3 \cos 2 x) d x

+ \frac{1}{64} \int_{0}^{π / 4} {(1 + \cos 4 x)}^{2} d x

= \frac{π}{64} + \frac{1}{8} + \frac{3 π}{64} + 0 - \frac{1}{96} + \frac{3}{32} + \frac{π}{256}

+ 0 + \frac{1}{128} \int_{0}^{π / 4} (1 + \cos 8 x) d x

= \frac{17 π}{256} + \frac{20}{96} + \frac{π}{512}

\Rightarrow A = \frac{35 π}{512} + \frac{20}{96}

________________________

B = \int_{0}^{π / 4} \sin^{8} x d x

= \frac{1}{16} \int_{0}^{π / 4} {(1 - \cos 2 x)}^{4} d x

= \frac{1}{16} \int_{0}^{π / 4} (1 - 4 \cos 2 x + 6 \cos^{2} 2 x

- 4 \cos^{3} 2 x + \cos^{4} 2 x) d x

= \frac{π}{64} - \frac{1}{8} + \frac{3 π}{64} + \frac{1}{96} - \frac{3}{32} + \frac{π}{256} + \frac{π}{512}

\Rightarrow B = \frac{35 π}{512} - \frac{20}{96}

________________________

A + B = \frac{35 π}{256}

A - B = \frac{5}{12}

A^{2} - B^{2} = \frac{175 π}{3072} .

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Aug/18

m e t h o d 1

{c o s}^{8} x + {s i n}^{8} x

= {({c o s}^{4} x + {s i n}^{4} x)}^{2} - 2 {c o s}^{4} {x s i n}^{4} x

= {{({c o s}^{2} x + {s i n}^{2} x)}^{2} - 2 {c o s}^{2} x . {s i n}^{2} x}^{2} - \frac{1}{8} \times {(2 s i n x c o s x)}^{4}

= {1 - \frac{1}{2} {(2 s i n x c o s x)}^{2}}^{2} - \frac{1}{8} \times {(s i n 2 x)}^{4}

= {\frac{2 - {s i n}^{2} 2 x}{2}}^{2} - \frac{1}{8} {(\frac{1 - c o s 4 x}{2})}^{2}

= {\frac{2 - \frac{1 - c o s 4 x}{2}}{2}}^{2} - \frac{1}{8} (\frac{1 - 2 c o s 4 x + \frac{1 + c o s 8 x}{2}}{4})

= {\frac{4 - 1 + c o s 4 x}{4}}^{2} - \frac{1}{8} (\frac{2 - 4 c o s 4 x + 1 + c o s 8 x}{8})

= {\frac{9 + 6 c o s 4 x + \frac{1 + c o s 8 x}{2}}{16}} - (\frac{c o s 8 x - 4 c o s 4 x + 3}{64})

= {\frac{18 + 12 c o s 4 x + 1 + c o s 8 x}{32}} - (\frac{c o s 8 x - 4 c o s 4 x + 3}{64})

= (\frac{c o s 8 x + 12 c o s 4 x + 19}{32}) - (\frac{c o s 8 x - 4 c o s 4 x + 3}{64})

= (\frac{2 c o s 8 x + 24 c o s 4 x + 38 - c o s 8 x + 4 c o s 4 x - 3}{64})

= (\frac{c o s 8 x + 28 c o s 4 x + 35}{64})

s o A + B =

\int_{0}^{\frac{Π}{4}} \frac{c o s 8 x + 28 c o s 4 x + 35}{64} d x

= \frac{1}{64} ∣ (\frac{s i n 8 x}{8} + \frac{28 s i n 4 x}{4} + 35 x) ∣_{0}^{\frac{Π}{4}}

= \frac{1}{64} (\frac{s i n 2 Π}{8} + \frac{28 s i n Π}{4} + \frac{35 \times \frac{Π}{4}}{})

= \frac{1}{64} \times \frac{35 Π}{4} = \frac{35 Π}{256} p l s c h e c k

i s h a l l s o l v e i t b y g a m m a b e t a f u n c t i o n l a t e r