let-I-0-pi-2-cos-6-x-dx-and-J-0-pi-2-sin-6-xdx-1-cslculate-I-J-and-I-J-2-find-the-value-of-I-and-J-

Question Number 41053 by turbo msup by abdo last updated on 01/Aug/18

l e t I = \int_{0}^{\frac{π}{2}} {c o s}^{6} x d x a n d

J = \int_{0}^{\frac{π}{2}} {s i n}^{6} x d x

1) c s 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 prof Abdo imad last updated on 01/Aug/18

2) w e h a v e I + J = \frac{5 π}{16} a n d I - J = 0 \Rightarrow

I = J a n d 2 I = \frac{5 π}{16} \Rightarrow I = J = \frac{5 π}{32} .

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Aug/18

I + J = \int_{0}^{\frac{Π}{2}} ({c o s}^{6} x + {s i n}^{6} x) d x

= \int_{0}^{\frac{Π}{2}} {({c o s}^{2} x + {s i n}^{2} x)}^{3} - 3 {c o s}^{2} {x s i n}^{2} x ({c o s}^{2} x + {s i n}^{2} x) d x

= \int_{0}^{\frac{Π}{2}} {1 - 3 (\frac{1 + c o s 2 x}{2}) (\frac{1 - c o s 2 x}{2})} d x

= \int_{0}^{\frac{Π}{2}} d x - \frac{3}{4} \int_{0}^{\frac{Π}{2}} (1 - {c o s}^{2} 2 x) d x

= \int_{0}^{\frac{Π}{2}} d x - \frac{3}{4} \int_{0}^{\frac{Π}{2}} d x + \frac{3}{4} \int_{0}^{\frac{Π}{2}} \frac{1 + c o s 4 x}{2} d x

= \int_{0}^{\frac{Π}{2}} d x - \frac{3}{4} \int_{0}^{\frac{Π}{2}} d x + \frac{3}{8} \int_{0}^{\frac{Π}{2}} d x + \frac{3}{8} \int_{0}^{\frac{Π}{2}} c o s 4 x d x

=∣ x - \frac{3}{4} x + \frac{3}{8} x ∣_{0}^{\frac{Π}{2}} + \frac{3}{8} \times ∣ \frac{s i n 4 x}{4} ∣_{0}^{\frac{Π}{2}}

= (1 - \frac{3}{4} + \frac{3}{8}) (\frac{Π}{2}) + 0

= (\frac{8 - 6 + 3}{8}) (\frac{Π}{2}) = \frac{5 Π}{16}

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Aug/18

I - J = \int_{0}^{\frac{Π}{2}} ({c o s}^{6} x - {s i n}^{6} x) d x

= \int_{0}^{\frac{Π}{2}} {({c o s}^{2} x - {s i n}^{2} x)}^{3} + 3 {c o s}^{2} {x s i n}^{2} x ({c o s}^{2} x - {s i n}^{2} x) d x

= \int_{0}^{\frac{Π}{2}} {c o s}^{3} 2 x + 3 (\frac{1 + c o s 2 x}{2}) (\frac{1 - c o s 2 x}{2}) (c o s 2 x) d x

= \int_{0}^{\frac{Π}{2}} \frac{c o s 6 x + 3 c o s 2 x}{4} + \frac{3}{4} (1 - \frac{1 + c o s 4 x}{2}) (c o s 2 x) d x

= \frac{1}{4} \int_{0}^{\frac{Π}{2}} c o s 6 x + \frac{3}{4} \int_{0}^{\frac{Π}{2}} c o s 2 x + \int_{0}^{\frac{Π}{2}} \frac{2 - 1 - c o s 4 x}{2} c o s 2 x d x

= 0 + 0 + \frac{1}{2} \int_{0}^{\frac{Π}{2}} \frac{(1 - c o s 4 x) c o s 2 x}{}

= \frac{1}{2} \int_{0}^{\frac{Π}{2}} (c o s 2 x) - \frac{1}{4} \int_{0}^{\frac{Π}{2}} c o s 6 x + c o s 2 x d x

= 0

\int_{0}^{\frac{Π}{2}} c o s 2 x d x =∣ \frac{s i n 2 x}{2} ∣_{0}^{\frac{Π}{2}} = 0

\int_{0}^{\frac{Π}{2}} c o s 6 x =∣ \frac{s i n 6 x}{6} ∣_{0}^{\frac{Π}{2}} = \frac{s i n 3 Π - s i n 0}{6} = 0

Commented by tanmay.chaudhury50@gmail.com last updated on 01/Aug/18

s o I = \frac{5 Π}{32} J = \frac{5 Π}{32}

Commented by prof Abdo imad last updated on 01/Aug/18

y o u r a n s w e r i s c o r r e c t s i r T a n m a y t h a n k s .

Answered by prof Abdo imad last updated on 01/Aug/18

1) w e u s e t h e f o r m u l a e a^{3} + b^{3} = {(a + b)}^{3} - 3 a b

a n d a^{3} - b^{3} = {(a - b)}^{3} + 3 a b

I + J = \int_{0}^{\frac{π}{2}} ({({c o s}^{2} x)}^{3} + {({s i n}^{2} x)}^{3}) d x

= \int_{0}^{\frac{π}{2}} {({c o s}^{2} x + {s i n}^{2} x)}^{3} - 3 {c o s}^{2} x {s i n}^{2} x) d x

= \frac{π}{2} - 3 \int_{0}^{\frac{π}{2}} \frac{1 + c o s (2 x)}{2} \frac{1 - c o s (2 x)}{2} d x

= \frac{π}{2} - \frac{3}{4} \int_{0}^{\frac{π}{2}} (1 - {c o s}^{2} (2 x)) d x

= \frac{π}{2} - \frac{3}{4} \int_{0}^{\frac{π}{2}} (1 - \frac{1 + c o s (4 x)}{2}) d x

= \frac{π}{2} - \frac{3}{8} \int_{0}^{\frac{π}{2}} (1 - c o s (4 x)) d x

= \frac{π}{2} - \frac{3 π}{16} + \frac{3}{8} \int_{0}^{\frac{π}{2}} c o s (4 x) d x

= \frac{5 π}{16} + \frac{3}{64} {[s i n (4 x)]}_{0}^{\frac{π}{2}} = \frac{5 π}{16}

Commented by tanmay.chaudhury50@gmail.com last updated on 01/Aug/18

a^{3} + b^{3} = {(a + b)}^{3} - 3 a b (a + b)

o r a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})

a^{3} - b^{3} = {(a - b)}^{3} + 3 a b (a - b)

o r a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})

Commented by math khazana by abdo last updated on 01/Aug/18

e r r o r a t t h e f i r s t l i n e

a^{3} + b^{3} = {(a + b)}^{3} - 3 a b (a + b) a n d

a^{3} - b^{3} = {(a - b)}^{3} + 3 a b (a - b)

Answered by prof Abdo imad last updated on 01/Aug/18

I - J = \int_{0}^{\frac{π}{2}} ({({c o s}^{2} x)}^{3} - {({s i n}^{2} x)}^{3}) d x

= \int_{0}^{\frac{π}{2}} {({c o s}^{2} x - {s i n}^{2} x)}^{3} + 3 {c o s}^{2} x {s i n}^{2} x ({c o s}^{2} x - {s i n}^{2} x) d x

= \int_{0}^{\frac{π}{2}} ({c o s}^{2} x - {s i n}^{2} x) {{({c o s}^{2} x - {s i n}^{2} x)}^{2} + 3 {c o s}^{2} {x s i n}^{2} x} d x

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

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

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

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

= \frac{1}{8} \int_{0}^{\frac{π}{2}} c o s (2 x) (7 - c o s (4 x)) d x

= \frac{7}{8} {[\frac{1}{2} s i n (2 x)]}_{0}^{\frac{π}{2}} - \frac{1}{8} \int_{0}^{\frac{π}{2}} c o s (2 x) c o s (4 x)

= - \frac{1}{16} \int_{0}^{\frac{π}{2}} {c o s (6 x) + c o s (2 x)) d x = 0 \Rightarrow I - J = 0