find A_n = ∫_0 ^(π/4) cos^n xdx and B_n =∫_0 ^(π/4) sin^n xdx 2) find ∫_0 ^(π/4) cos^6 xdx and ∫


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Question Number 48498 by maxmathsup by imad last updated on 24/Nov/18

$f i n d A_{n} = \int_{0}^{\frac{π}{4}} {c o s}^{n} x d x a n d B_{n} = \int_{0}^{\frac{π}{4}} {s i n}^{n} x d x$ $2) f i n d \int_{0}^{\frac{π}{4}} {c o s}^{6} x d x a n d \int_{0}^{\frac{π}{4}} {s i n}^{6} x d x .$
Commented by Abdulhafeez Abu qatada last updated on 24/Nov/18

$i m a d e a n e r r o r, b u t i t s c o r r e c t e d$
Commented by maxmathsup by imad last updated on 25/Nov/18
$1) W_n =∫_0 ^(π/4) (((e^(ix) +e^(−ix) )/2))^n dx =(1/2^n ) ∫_0 ^(π/4) (e^(ix) +e^(−ix) )^n dx =(1/2^n ) ∫_0 ^(π/4) (Σ_(k=0) ^n C_n ^k (e^(ix) )^k (e^(−ix) )^(n−k) )dx =(1/2^n ) Σ_(k=0) ^n C_n ^k ∫_0 ^(π/4) e^(ikx) e^(−i(n−k)x) dx =(1/2^n ) Σ_(k=0) ^n C_n ^k A_k A_k =∫_0 ^(π/4) e^((ik−in +ik)x) dx =∫_0 ^(π/4) e^((2ik−in)x) dx =∫_0 ^(π/4) e^(i(2k−n)x) dx =∫_0 ^(π/4) cos(2k−n)x dx +i ∫_0 ^(π/4) sin(2k−n)x dx =[(1/(2k−n))sin(2k−n)x]_0 ^(π/4) −i [(1/(2k−n)) cos(2k−n)x]_0 ^(π/4) =(1/(2k−n))sin(2k−n)(π/4) +i ((1/(2k−n)) −(1/(2k−n)) cos(2k−n)(π/4)) ⇒ W_n =(1/2^n ) Σ_(k=0) ^n C_n ^k (1/(2k−n)){sin(2k−n)(π/4) +i(1−cos(2k−n)(π/4))} =(1/2^n ) Σ_(k=0) ^n C_n ^k ((sin(2k−n)(π/4))/(2k−n)) +i (1/2^n ) Σ_(k=0) ^n C_n ^k ((1−cos(2k−n)(π/4))/(2k−n)) but W_n is real ⇒ (1/2^n ) Σ_(k=0) ^n ((1−cos(2k−n)(π/4))/(2k−n)) =0 and W_n =(1/2^n ) Σ_(k=0) ^n C_n ^k ((sin(2k−n)(π/4))/(2k−n)) .$
$1) W_{n} = \int_{0}^{\frac{π}{4}} {(\frac{e^{i x} + e^{- i x}}{2})}^{n} d x = \frac{1}{2^{n}} \int_{0}^{\frac{π}{4}} {(e^{i x} + e^{- i x})}^{n} d x$ $= \frac{1}{2^{n}} \int_{0}^{\frac{π}{4}} (\sum_{k = 0}^{n} C_{n}^{k} {(e^{i x})}^{k} {(e^{- i x})}^{n - k}) d x$ $= \frac{1}{2^{n}} \sum_{k = 0}^{n} C_{n}^{k} \int_{0}^{\frac{π}{4}} e^{i k x} e^{- i (n - k) x} d x = \frac{1}{2^{n}} \sum_{k = 0}^{n} C_{n}^{k} A_{k}$ $A_{k} = \int_{0}^{\frac{π}{4}} e^{(i k - i n + i k) x} d x = \int_{0}^{\frac{π}{4}} e^{(2 i k - i n) x} d x = \int_{0}^{\frac{π}{4}} e^{i (2 k - n) x} d x$ $= \int_{0}^{\frac{π}{4}} c o s (2 k - n) x d x + i \int_{0}^{\frac{π}{4}} s i n (2 k - n) x d x$ $= {[\frac{1}{2 k - n} s i n (2 k - n) x]}_{0}^{\frac{π}{4}} - i {[\frac{1}{2 k - n} c o s (2 k - n) x]}_{0}^{\frac{π}{4}}$ $= \frac{1}{2 k - n} s i n (2 k - n) \frac{π}{4} + i (\frac{1}{2 k - n} - \frac{1}{2 k - n} c o s (2 k - n) \frac{π}{4}) \Rightarrow$ $W_{n} = \frac{1}{2^{n}} \sum_{k = 0}^{n} C_{n}^{k} \frac{1}{2 k - n} {s i n (2 k - n) \frac{π}{4} + i (1 - c o s (2 k - n) \frac{π}{4})}$ $= \frac{1}{2^{n}} \sum_{k = 0}^{n} C_{n}^{k} \frac{s i n (2 k - n) \frac{π}{4}}{2 k - n} + i \frac{1}{2^{n}} \sum_{k = 0}^{n} C_{n}^{k} \frac{1 - c o s (2 k - n) \frac{π}{4}}{2 k - n} b u t W_{n} i s r e a l \Rightarrow$ $\frac{1}{2^{n}} \sum_{k = 0}^{n} \frac{1 - c o s (2 k - n) \frac{π}{4}}{2 k - n} = 0 a n d W_{n} = \frac{1}{2^{n}} \sum_{k = 0}^{n} C_{n}^{k} \frac{s i n (2 k - n) \frac{π}{4}}{2 k - n} .$
Commented by maxmathsup by imad last updated on 25/Nov/18
$we have B_(n ) =∫_0 ^(π/4) (((e^(ix) −e^(−ix) )/(2i)))^n dx =(1/((2i)^n )) ∫_0 ^(π/4) (Σ_(k=0) ^n C_n ^k (e^(ix) )^k (−1)^(n−k) (e^(−ix) )^(n−k) dx =(1/((2i)^n )) Σ_(k=0) ^n C_n ^k (−1)^(n−k) ∫_0 ^(π/4) e^(ikx) e^(−i(n−k)x) dx =(1/((2i)^n )) Σ_(k=0) ^n (−1)^(n−k) C_n ^k .A_k with A_k =∫_0 ^(π/4) e^((ik−in+ik)x) dx =∫_0 ^(π/4) e^(i(2k−n)x) dx =∫_0 ^(π/4) cos(2k−n)x dx +i∫_0 ^(π/4) sin(2k−n)x dx =[(1/(2k−n))sin(2k−n)x]_0 ^(π/4) +i[−(1/(2k−n))cos(2k−n)x]_0 ^(π/4) =(1/(2k−n))sin(2k−n)(π/4) +(i/(2k−n))( 1−cos(2k−n)(π/4)) ⇒ B_n =(1/((2i)^n )) Σ_(k=0) ^n (−1)^(n−k) C_n ^k (1/(2k−n)){ sin(2k−n)(π/4) +i(1−cos(2k−n)(π/4))} ⇒B_(2n) = (1/(2^(2n) (−1)^n )) Σ_(k=0) ^(2n) (−1)^k C_(2n) ^k (1/(2k−2n)){sin(k−n)(π/2) +i(1−cos(k−n)(π/2))} but B_(2n) is real ⇒ B_(2n) =(((−1)^n )/2^(2n+1) ) Σ_(k=0) ^(2n) (−1)^k C_(2n) ^k ((sin(k−n)(π/2))/(k−n)) and we follow the same method to find B_(2n+1)$
$w e h a v e B_{n} = \int_{0}^{\frac{π}{4}} {(\frac{e^{i x} - e^{- i x}}{2 i})}^{n} d x = \frac{1}{{(2 i)}^{n}} \int_{0}^{\frac{π}{4}} (\sum_{k = 0}^{n} C_{n}^{k} {(e^{i x})}^{k} {(- 1)}^{n - k} {(e^{- i x})}^{n - k} d x$ $= \frac{1}{{(2 i)}^{n}} \sum_{k = 0}^{n} C_{n}^{k} {(- 1)}^{n - k} \int_{0}^{\frac{π}{4}} e^{i k x} e^{- i (n - k) x} d x$ $= \frac{1}{{(2 i)}^{n}} \sum_{k = 0}^{n} {(- 1)}^{n - k} C_{n}^{k} . A_{k} w i t h A_{k} = \int_{0}^{\frac{π}{4}} e^{(i k - i n + i k) x} d x$ $= \int_{0}^{\frac{π}{4}} e^{i (2 k - n) x} d x = \int_{0}^{\frac{π}{4}} c o s (2 k - n) x d x + i \int_{0}^{\frac{π}{4}} s i n (2 k - n) x d x$ $= {[\frac{1}{2 k - n} s i n (2 k - n) x]}_{0}^{\frac{π}{4}} + i {[- \frac{1}{2 k - n} c o s (2 k - n) x]}_{0}^{\frac{π}{4}}$ $= \frac{1}{2 k - n} s i n (2 k - n) \frac{π}{4} + \frac{i}{2 k - n} (1 - c o s (2 k - n) \frac{π}{4}) \Rightarrow$ $B_{n} = \frac{1}{{(2 i)}^{n}} \sum_{k = 0}^{n} {(- 1)}^{n - k} C_{n}^{k} \frac{1}{2 k - n} {s i n (2 k - n) \frac{π}{4} + i (1 - c o s (2 k - n) \frac{π}{4})}$ $\Rightarrow B_{2 n} = \frac{1}{2^{2 n} {(- 1)}^{n}} \sum_{k = 0}^{2 n} {(- 1)}^{k} C_{2 n}^{k} \frac{1}{2 k - 2 n} {s i n (k - n) \frac{π}{2} + i (1 - c o s (k - n) \frac{π}{2})}$ $b u t B_{2 n} i s r e a l \Rightarrow B_{2 n} = \frac{{(- 1)}^{n}}{2^{2 n + 1}} \sum_{k = 0}^{2 n} {(- 1)}^{k} C_{2 n}^{k} \frac{s i n (k - n) \frac{π}{2}}{k - n}$ $a n d w e f o l l o w t h e s a m e m e t h o d t o f i n d B_{2 n + 1}$
Commented by maxmathsup by imad last updated on 25/Nov/18
$2) let I =∫_0 ^(π/4) cos^6 x dx and J =∫_0 ^(π/4) sin^6 x dx ⇒ I +J =∫_0 ^(π/4) { (cos^2 x)^(3 ) +(sin^2 x)^3 }dx =∫_0 ^(π/4) { (cos^2 x+sin^2 x)^3 −3cos^2 x sin^2 x(cos^2 x +sin^2 x)}dx =∫_0 ^(π/4) {1−3 cos^2 x sin^2 x}dx =∫_0 ^(π/4) {1−(3/4)(1+cos(2x))(1−cos(2x))dx =(π/4) −(3/4) ∫_0 ^(π/4) (1−cos^2 (2x))dx =(π/4) −(3/4).(π/4) +(3/4) ∫_0 ^(π/4) ((1+cos(4x))/2)dx =(π/(16)) +(3/8).(π/4) +(3/8) ∫_0 ^(π/4) cos(4x)dx =(π/(16)) +((3π)/(32)) +0 =((5π)/(32)) also we have I−J = ∫_0 ^(π/4) ( (cos^2 x)^3 −(sin^2 x)^3 dx =∫_0 ^(π/4) (cos^2 x −sin^2 x)(cos^4 x +cos^2 x sin^2 x +sin^4 x)dx =∫_0 ^(π/4) cos(2x)( (cos^2 x +sin^2 x)^2 −cos^2 x sin^2 x)dx =∫_0 ^(π/4) cos(2x)(1−(1/4)sin^2 (2x))dx=∫_0 ^(π/4) cos(2x)(1−(1/8)(1−cos(4x))dx =(7/8) ∫_0 ^(π/4) cos(2x)dx +(1/8) ∫_0 ^(π/4) cos(2x)cos(4x)dx but ∫_0 ^(π/4) cos(2x)dx =[(1/2)sin(2x)]_0 ^(π/4) =(1/2) ∫_0 ^(π/4) cos(2x)cos(4x) dx =(1/2) ∫_0 ^(π/4) (cos(6x)+cos(2x))dx =(1/(12))[sin(6x)]_0 ^(π/4) +(1/4)[sin(2x)]_0 ^(π/4) =(1/(12))sin(((3π)/2)) +(1/4) =−(1/(12)) +(1/4) =((−1+3)/(12)) =(1/6) ⇒ I −J =(7/(16)) +(1/(48)) =((22)/(48)) =((11)/(24)) ⇒ I +J =((5π)/(32)) and I −J =((11)/(24)) ⇒ 2I =((5π)/(32)) +((11)/(24)) ⇒ I =((5π)/(64)) +((11)/(48)) also 2J =((5π)/(32)) −((11)/(24)) ⇒ J =((5π)/(64)) −((11)/(48)) . J$
$2) l e t I = \int_{0}^{\frac{π}{4}} {c o s}^{6} x d x a n d J = \int_{0}^{\frac{π}{4}} {s i n}^{6} x d x \Rightarrow$ $I + J = \int_{0}^{\frac{π}{4}} {{({c o s}^{2} x)}^{3} + {({s i n}^{2} x)}^{3}} d x$ $= \int_{0}^{\frac{π}{4}} {{({c o s}^{2} x + {s i n}^{2} x)}^{3} - 3 {c o s}^{2} x {s i n}^{2} x ({c o s}^{2} x + {s i n}^{2} x)} d x$ $= \int_{0}^{\frac{π}{4}} {1 - 3 {c o s}^{2} x {s i n}^{2} x} d x = \int_{0}^{\frac{π}{4}} {1 - \frac{3}{4} (1 + c o s (2 x)) (1 - c o s (2 x)) d x$ $= \frac{π}{4} - \frac{3}{4} \int_{0}^{\frac{π}{4}} (1 - {c o s}^{2} (2 x)) d x$ $= \frac{π}{4} - \frac{3}{4} . \frac{π}{4} + \frac{3}{4} \int_{0}^{\frac{π}{4}} \frac{1 + c o s (4 x)}{2} d x$ $= \frac{π}{16} + \frac{3}{8} . \frac{π}{4} + \frac{3}{8} \int_{0}^{\frac{π}{4}} c o s (4 x) d x$ $= \frac{π}{16} + \frac{3 π}{32} + 0 = \frac{5 π}{32} a l s o w e h a v e I - J = \int_{0}^{\frac{π}{4}} ({({c o s}^{2} x)}^{3} - {({s i n}^{2} x)}^{3} d x$ $= \int_{0}^{\frac{π}{4}} ({c o s}^{2} x - {s i n}^{2} x) ({c o s}^{4} x + {c o s}^{2} x {s i n}^{2} x + {s i n}^{4} x) d x$ $= \int_{0}^{\frac{π}{4}} c o s (2 x) ({({c o s}^{2} x + {s i n}^{2} x)}^{2} - {c o s}^{2} x {s i n}^{2} x) d x$ $= \int_{0}^{\frac{π}{4}} c o s (2 x) (1 - \frac{1}{4} {s i n}^{2} (2 x)) d x = \int_{0}^{\frac{π}{4}} c o s (2 x) (1 - \frac{1}{8} (1 - c o s (4 x)) d x$ $= \frac{7}{8} \int_{0}^{\frac{π}{4}} c o s (2 x) d x + \frac{1}{8} \int_{0}^{\frac{π}{4}} c o s (2 x) c o s (4 x) d x b u t$ $\int_{0}^{\frac{π}{4}} c o s (2 x) d x = {[\frac{1}{2} s i n (2 x)]}_{0}^{\frac{π}{4}} = \frac{1}{2}$ $\int_{0}^{\frac{π}{4}} c o s (2 x) c o s (4 x) d x = \frac{1}{2} \int_{0}^{\frac{π}{4}} (c o s (6 x) + c o s (2 x)) d x$ $= \frac{1}{12} {[s i n (6 x)]}_{0}^{\frac{π}{4}} + \frac{1}{4} {[s i n (2 x)]}_{0}^{\frac{π}{4}} = \frac{1}{12} s i n (\frac{3 π}{2}) + \frac{1}{4} = - \frac{1}{12} + \frac{1}{4} = \frac{- 1 + 3}{12} = \frac{1}{6}$ $\Rightarrow I - J = \frac{7}{16} + \frac{1}{48} = \frac{22}{48} = \frac{11}{24} \Rightarrow I + J = \frac{5 π}{32} a n d I - J = \frac{11}{24} \Rightarrow$ $2 I = \frac{5 π}{32} + \frac{11}{24} \Rightarrow I = \frac{5 π}{64} + \frac{11}{48} a l s o 2 J = \frac{5 π}{32} - \frac{11}{24} \Rightarrow J = \frac{5 π}{64} - \frac{11}{48} .$ $J$
	Answered by Abdulhafeez Abu qatada last updated on 24/Nov/18

	$I f I_{n} = \int {s i n}^{n} x d x, I_{n} = \frac{- \cos x . \sin^{n - 1} x}{n} + \frac{n - 1}{n} I_{n - 2}$ $∴ B_{n} = \underset{0}{\int^{\frac{π}{4}}} {s i n}^{n} x d x, B_{n} = {[\frac{- \cos x . \sin^{n - 1} x}{n} + \frac{n - 1}{n} A_{n - 2}]}_{0}^{\frac{π}{4}}$ $B_{n} = \frac{- {(\frac{\sqrt{2}}{2})}^{n}}{n} + \frac{n - 1}{n} B_{n - 2}$ $A_{n} = {[\frac{\sin x . \cos^{n - 1} x}{n} + \frac{n - 1}{n} A_{n - 2}]}_{0}^{\frac{π}{4}}$ $A_{n} = \frac{{(\frac{\sqrt{2}}{2})}^{n}}{n} + \frac{n - 1}{n} A_{n - 2}$ $A_{6} = \frac{{(\frac{\sqrt{2}}{2})}^{6}}{6} + \frac{5}{6} A_{4}$ $A_{6} = \frac{{(\frac{\sqrt{2}}{2})}^{6}}{6} + \frac{5}{6} (\frac{{(\frac{\sqrt{2}}{2})}^{4}}{4} + \frac{3}{4} A_{2})$ $A_{6} = \frac{{(\frac{\sqrt{2}}{2})}^{6}}{6} + \frac{5}{6} (\frac{{(\frac{\sqrt{2}}{2})}^{4}}{4}) + \frac{5}{6} . \frac{3}{4} (\frac{{(\frac{\sqrt{2}}{2})}^{2}}{2} + \frac{1}{2} A_{0})$ $A_{0} = {\int_{0}}^{\frac{π}{4}} d x = \frac{π}{4}$ $A_{6} = \frac{{(\frac{\sqrt{2}}{2})}^{6}}{6} + \frac{5}{6} (\frac{{(\frac{\sqrt{2}}{2})}^{4}}{4}) + \frac{5}{6} . \frac{3}{4} (\frac{{(\frac{\sqrt{2}}{2})}^{2}}{2} + \frac{π}{8})$ $A_{6} = \frac{5 π}{64} + \frac{11}{48} . . . . . A f t e r s i m p l i f y i n g$ $B_{6} = \frac{- {(\frac{\sqrt{2}}{2})}^{6}}{6} + \frac{5}{6} B_{4}$ $B_{6} = \frac{- {(\frac{\sqrt{2}}{2})}^{6}}{6} + \frac{5}{6} (\frac{- {(\frac{\sqrt{2}}{2})}^{4}}{4} + \frac{3}{4} B_{2})$ $B_{6} = \frac{- {(\frac{\sqrt{2}}{2})}^{6}}{6} + \frac{5}{6} (\frac{- {(\frac{\sqrt{2}}{2})}^{4}}{4}) + \frac{5}{6} . \frac{3}{4} (\frac{- {(\frac{\sqrt{2}}{2})}^{2}}{2} + \frac{1}{2} B_{0})$ $B_{0} = {\int_{0}}^{\frac{π}{4}} d x = \frac{π}{4}$ $B_{6} = \frac{- {(\frac{\sqrt{2}}{2})}^{6}}{6} + \frac{5}{6} (\frac{- {(\frac{\sqrt{2}}{2})}^{4}}{4}) + \frac{5}{6} . \frac{3}{4} (\frac{- {(\frac{\sqrt{2}}{2})}^{2}}{2} + \frac{π}{8})$ $B_{6} = \frac{5 π}{64} - \frac{11}{48} . . . . . . A f t e r s i m p l i f y i n g$ $A b u q a t a d a$
	Answered by Abdulhafeez Abu qatada last updated on 24/Nov/18

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