calculate A = ∫_0 ^(π/4) cos^8 xdx and B= ∫_0 ^(π/4) sin^8 xdx 2) calculate A +B and A−B 3) calculate A^2 −B^2


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 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
$let A_n = ∫_0 ^(π/4) cos^(2n) xdx we haveA =A_4 A_(n+1) = ∫_0 ^(π/4) cos^(2n) (1−sin^2 x)dx =A_n −∫_0 ^(π/4) sin^2 x cos^(2n) xdx by parts ∫_0 ^(π/4) sinx(−sinx)cos^(2n) x dx = [(1/(2n+1))sinxcos^(2n+1) x ]_0 ^(π/4) −∫_0 ^(π/4) (1/(2n+1)) cos^(2n+2) xdx =(1/(2n+1))( (1/(√2)))^(2n+2) −(1/(2n+1)) A_(n+1) ⇒ (1+(1/(2n+1)))A_(n+1) =A_n +(1/((2n+1)2^(n+1) )) ⇒ ((2n+2)/(2n+1)) A_(n+1) = A_n +(1/((2n+1)2^(n+1) )) ⇒ A_(n+1) =((2n+1)/(2n+2)) A_n + (1/((2n+2)2^(n+1) )) ⇒ A_4 =(7/8) A_3 + (1/(8.2^4 )) A_3 =(5/6) A_2 + (1/(6.2^3 )) A_2 = (3/4) A_1 + (1/(4.2^2 )) A_1 =(1/2) A_0 +(1/4) ⇒ A_4 =(7/8){(5/6) A_2 +(1/(6.2^3 ))} +(1/(8.2^4 )) = ((35)/(48)) A_2 +(7/(48.2^3 )) +(1/(8.2^4 )) =((35)/(48)){ (3/4) A_1 +(1/(4.2^2 ))} +(7/(48.2^3 )) +(1/(8.2^4 )) = ((105)/(192)) A_1 + ((35)/(192.2^2 )) +(7/(48.2^3 )) +(1/(8.2^4 )) =((105)/(192)){(1/2)A_0 +(1/4)} +((35)/(192.2^2 )) +(7/(48.2^3 )) +(1/(8.2^4 )) =((105)/(192)){ (π/8) +(1/4)} +((35)/(192.2^2 )) +(7/(48.2^3 )) +(1/(8.2^4 )) =A$
$l e t A_{n} = \int_{0}^{\frac{π}{4}} {c o s}^{2 n} x d x w e h a v e A = A_{4}$ $A_{n + 1} = \int_{0}^{\frac{π}{4}} {c o s}^{2 n} (1 - {s i n}^{2} x) d x$ $= A_{n} - \int_{0}^{\frac{π}{4}} {s i n}^{2} x {c o s}^{2 n} x d x b y p a r t s$ $\int_{0}^{\frac{π}{4}} s i n x (- s i n x) {c o s}^{2 n} x d x$ $= {[\frac{1}{2 n + 1} {s i n x c o s}^{2 n + 1} x]}_{0}^{\frac{π}{4}} - \int_{0}^{\frac{π}{4}} \frac{1}{2 n + 1} {c o s}^{2 n + 2} x d x$ $= \frac{1}{2 n + 1} {(\frac{1}{\sqrt{2}})}^{2 n + 2} - \frac{1}{2 n + 1} A_{n + 1} \Rightarrow$ $(1 + \frac{1}{2 n + 1}) A_{n + 1} = A_{n} + \frac{1}{(2 n + 1) 2^{n + 1}} \Rightarrow$ $\frac{2 n + 2}{2 n + 1} A_{n + 1} = A_{n} + \frac{1}{(2 n + 1) 2^{n + 1}} \Rightarrow$ $A_{n + 1} = \frac{2 n + 1}{2 n + 2} A_{n} + \frac{1}{(2 n + 2) 2^{n + 1}} \Rightarrow$ $A_{4} = \frac{7}{8} A_{3} + \frac{1}{8 . 2^{4}}$ $A_{3} = \frac{5}{6} A_{2} + \frac{1}{6 . 2^{3}}$ $A_{2} = \frac{3}{4} A_{1} + \frac{1}{4 . 2^{2}}$ $A_{1} = \frac{1}{2} A_{0} + \frac{1}{4} \Rightarrow$ $A_{4} = \frac{7}{8} {\frac{5}{6} A_{2} + \frac{1}{6 . 2^{3}}} + \frac{1}{8 . 2^{4}}$ $= \frac{35}{48} A_{2} + \frac{7}{48 . 2^{3}} + \frac{1}{8 . 2^{4}}$ $= \frac{35}{48} {\frac{3}{4} A_{1} + \frac{1}{4 . 2^{2}}} + \frac{7}{48 . 2^{3}} + \frac{1}{8 . 2^{4}}$ $= \frac{105}{192} A_{1} + \frac{35}{192 . 2^{2}} + \frac{7}{48 . 2^{3}} + \frac{1}{8 . 2^{4}}$ $= \frac{105}{192} {\frac{1}{2} A_{0} + \frac{1}{4}} + \frac{35}{192 . 2^{2}} + \frac{7}{48 . 2^{3}} + \frac{1}{8 . 2^{4}}$ $= \frac{105}{192} {\frac{π}{8} + \frac{1}{4}} + \frac{35}{192 . 2^{2}} + \frac{7}{48 . 2^{3}} + \frac{1}{8 . 2^{4}} = A$
Commented by math khazana by abdo last updated on 12/Aug/18

$l e t B_{n} = \int_{0}^{\frac{π}{4}} {s i n}^{2 n} x d x w e h a v e B = B_{4}$ $B_{n + 1} = \int_{0}^{\frac{π}{4}} {s i n}^{2 n} (1 - {c o s}^{2} x) d x$ $= \int_{0}^{\frac{π}{4}} {s i n}^{2 n} x d x - \int_{0}^{\frac{π}{4}} c o s x (c o s x) {s i n}^{2 n} x d x b y$ $p a r t s$ $\int_{0}^{\frac{π}{4}} c o s x (c o s x) {s i n}^{2 n} x d x = {[\frac{1}{2 n + 1} {c o s x s i n}^{2 n + 1} x]}_{0}^{\frac{π}{4}}$ $- \int_{0}^{\frac{π}{4}} (- s i n x) \frac{1}{2 n + 1} {s i n}^{2 n + 1} x d x$ $= \frac{1}{(2 n + 1)} {(\frac{1}{\sqrt{2}})}^{2 n + 2} + \frac{1}{2 n + 1} \int_{0}^{\frac{π}{4}} {s i n}^{2 n + 2} x d x$ $= \frac{1}{(2 n + 1) 2^{n + 1}} + \frac{1}{2 n + 1} B_{n + 1} \Rightarrow$ $B_{n + 1} = B_{n} - \frac{1}{(2 n + 1) 2^{n + 1}} - \frac{1}{2 n + 1} B_{n + 1} \Rightarrow$ $(1 + \frac{1}{2 n + 1}) B_{n + 1} = B_{n} - \frac{1}{(2 n + 1) 2^{n + 1}} \Rightarrow$ $\frac{2 n + 2}{2 n + 1} B_{n + 1} = B_{n} - \frac{1}{(2 n + 1) 2^{n + 1}} \Rightarrow$ $B_{n + 1} = \frac{2 n + 1}{2 n + 2} B_{n} - \frac{1}{(2 n + 2) 2^{n + 1}} . . . b e c o n t i n u e d . . .$
	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
	$method 1 cos^8 x+sin^8 x =(cos^4 x+sin^4 x)^2 −2cos^4 xsin^4 x ={(cos^2 x+sin^2 x)^2 −2cos^2 x.sin^2 x}^2 −(1/8)×(2sinxcosx)^4 ={1−(1/2)(2sinxcosx)^2 }^2 −(1/8)×(sin2x)^4 ={((2−sin^2 2x)/2)}^2 −(1/8)(((1−cos4x)/2))^2 ={((2−((1−cos4x)/2))/2)}^2 −(1/8)(((1−2cos4x+((1+cos8x)/2))/4)) ={((4−1+cos4x)/4)}^2 −(1/8)(((2−4cos4x+1+cos8x)/8)) ={((9+6cos4x+((1+cos8x)/2))/(16))}−(((cos8x−4cos4x+3)/(64))) ={((18+12cos4x+1+cos8x)/(32))}−(((cos8x−4cos4x+3)/(64))) =(((cos8x+12cos4x+19)/(32)))−(((cos8x−4cos4x+3)/(64))) =(((2cos8x+24cos4x+38−cos8x+4cos4x−3)/(64))) =(((cos8x+28cos4x+35)/(64))) so A+B= ∫_0 ^(Π/4) ((cos8x+28cos4x+35)/(64))dx =(1/(64))∣(((sin8x)/8)+((28sin4x)/4)+35x)∣_0 ^(Π/4) =(1/(64))(((sin2Π)/8)+((28sinΠ)/4)+((35×(Π/4))/)) =(1/(64))×((35Π)/4)=((35Π)/(256)) pls check i shall solve it by gamma beta function later$
	$m e t h o d 1$ ${c o s}^{8} x + {s i n}^{8} x$ $= {({c o s}^{4} x + {s i n}^{4} x)}^{2} - 2 {c o s}^{4} {x s i n}^{4} x$ $= {{({c o s}^{2} x + {s i n}^{2} x)}^{2} - 2 {c o s}^{2} x . {s i n}^{2} x}^{2} - \frac{1}{8} \times {(2 s i n x c o s x)}^{4}$ $= {1 - \frac{1}{2} {(2 s i n x c o s x)}^{2}}^{2} - \frac{1}{8} \times {(s i n 2 x)}^{4}$ $= {\frac{2 - {s i n}^{2} 2 x}{2}}^{2} - \frac{1}{8} {(\frac{1 - c o s 4 x}{2})}^{2}$ $= {\frac{2 - \frac{1 - c o s 4 x}{2}}{2}}^{2} - \frac{1}{8} (\frac{1 - 2 c o s 4 x + \frac{1 + c o s 8 x}{2}}{4})$ $= {\frac{4 - 1 + c o s 4 x}{4}}^{2} - \frac{1}{8} (\frac{2 - 4 c o s 4 x + 1 + c o s 8 x}{8})$ $= {\frac{9 + 6 c o s 4 x + \frac{1 + c o s 8 x}{2}}{16}} - (\frac{c o s 8 x - 4 c o s 4 x + 3}{64})$ $= {\frac{18 + 12 c o s 4 x + 1 + c o s 8 x}{32}} - (\frac{c o s 8 x - 4 c o s 4 x + 3}{64})$ $= (\frac{c o s 8 x + 12 c o s 4 x + 19}{32}) - (\frac{c o s 8 x - 4 c o s 4 x + 3}{64})$ $= (\frac{2 c o s 8 x + 24 c o s 4 x + 38 - c o s 8 x + 4 c o s 4 x - 3}{64})$ $= (\frac{c o s 8 x + 28 c o s 4 x + 35}{64})$ $s o A + B =$ $\int_{0}^{\frac{Π}{4}} \frac{c o s 8 x + 28 c o s 4 x + 35}{64} d x$ $= \frac{1}{64} ∣ (\frac{s i n 8 x}{8} + \frac{28 s i n 4 x}{4} + 35 x) ∣_{0}^{\frac{Π}{4}}$ $= \frac{1}{64} (\frac{s i n 2 Π}{8} + \frac{28 s i n Π}{4} + \frac{35 \times \frac{Π}{4}}{})$ $= \frac{1}{64} \times \frac{35 Π}{4} = \frac{35 Π}{256} p l s c h e c k$ $i s h a l l s o l v e i t b y g a m m a b e t a f u n c t i o n l a t e r$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum