Evaluate : 1) ((∫_0 ^( 1_ ) (1−(1−x^2 )^(100) )^(201) .xdx)/(∫_0 ^( 1) (1−(1−x^2 )^(100) )^(202) .xdx)) = ? 2) ((∫_0 ^( 1) (1−x^(200) )^(201) dx)/(∫


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Question Number 56280 by rahul 19 last updated on 13/Mar/19

$E v a l u a t e :$ $1) \frac{\int_{0}^{1_{}} {(1 - {(1 - x^{2})}^{100})}^{201} . x d x}{\int_{0}^{1} {(1 - {(1 - x^{2})}^{100})}^{202} . x d x} = ?$ $2) \frac{\int_{0}^{1} {(1 - x^{200})}^{201} d x}{\int_{0}^{1} {(1 - x^{200})}^{202} d x} = ?$
Commented by rahul 19 last updated on 13/Mar/19

$A n s :$ $1) \frac{20201}{20200} 2) \frac{40401}{40400} .$
Commented by tanmay.chaudhury50@gmail.com last updated on 13/Mar/19

$r a h u l p l s c h e c k t h e a n s w e r$
	Answered by tanmay.chaudhury50@gmail.com last updated on 13/Mar/19

	$2) t = x^{200} d t = 200 x^{199} d x$ $d t = 200 \times {(t^{\frac{1}{200}})}^{199} d x$ $d x = \frac{d t}{200 \times t^{\frac{199}{200}}}$ $\frac{\int_{0}^{1} {(1 - x^{200})}^{201} d x}{\int_{0}^{1} {(1 - x^{200})}^{202} d x}$ $\frac{\int_{0}^{1} {(1 - t)}^{201} \times \frac{d t}{200 \times t^{\frac{199}{200}}}}{\int_{0}^{1} {(1 - t)}^{202} \times \frac{d t}{200 \times t^{\frac{199}{200}}}}$ $\frac{\int_{0}^{1} {(1 - t)}^{201} \times t^{\frac{1}{200} - 1} d t}{\int_{0}^{1} {(1 - t)}^{202} \times t^{\frac{1}{200} - 1} d t}$ $n o w l o o k f o r m u l a β f u n c t i o n$ $β (m, n) = \int_{0}^{1} x^{m - 1} {(1 - x)}^{n - 1} d x$ $= \frac{⌈ (m) ⌈ (n)}{⌈ (m + n)}$ $= \frac{\frac{⌈ (202) ⌈ (\frac{1}{200})}{⌈ (202 + \frac{1}{200})}}{\frac{⌈ (203) ⌈ (\frac{1}{200})}{⌈ (203 + \frac{1}{200})}}$ $n o w ⌈ (n + 1) = n!$ $= \frac{\frac{201!}{(201 + \frac{1}{200})!}}{\frac{202!}{(202 + \frac{1}{200})!}}$ $= \frac{201! \times (202 + \frac{1}{200})!}{202! \times (201 + \frac{1}{200})!}$ $= \frac{1}{202} \times (202 + \frac{1}{200}) \times (201 + \frac{1}{200})! \times \frac{1}{(201 + \frac{1}{200})!}$ $= 1 + \frac{1}{200 \times 202}$ $= l$
	Commented by rahul 19 last updated on 13/Mar/19

	$I t w i l l b e$ $= \frac{1}{202} \times (202 + \frac{1}{200}) = 1 + \frac{1}{202 \times 200}$ $= \frac{40401}{40400} A n s .!$ $T h a n k y o u S i r!$
	Commented by tanmay.chaudhury50@gmail.com last updated on 13/Mar/19

	$t h a n k y o u r a h u l . . . s o m e t i m e s i l o s s c o n s i s t s n c y . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 13/Mar/19
	$1)((∫_0 ^1 {1−(1−x^2 )^a }^(2a+1) .xdx)/(∫_0 ^1 {1−(1−x^2 )^a }^(2a+2) .xdx)) a=100 t=1−x^2 dt=−2xdx (((1/2)∫_1 ^0 (1−t^a )^(2a+1) ×−dt)/((1/2)∫_1 ^0 (1−t^a )^(2a+2) ×−dt)) ((∫_0 ^1 (1−t^a )^(2a+1) dt)/(∫_0 ^1 (1−t^a )^(2a+2) dt)) ((∫_0 ^1 (1−t^(100) )^(201) dt)/(∫_0 ^1 (1−t^(100) )^(202) dt)) look same type question of question no 2 but solving other way let t^(100) =sin^2 α →100t^(99) dt=2sinαcosαdα dt=((2sinαcosαdα)/(100(sin^2 α)^(1/(100)) ))=(1/(50))×(sinα)^(1−(1/(50))) ×cosαdα ((∫_0 ^(π/2) (1−sin^2 α)^(201) ×(1/(50))×(sinα)^(1−(1/(50))) ×cosαdα)/(∫_0 ^(π/2) (1−sin^2 α)^(202) ×(1/(50))×(sinα)^(1−(1/(50))) ×cosαdα))(/) ((∫^(π/2) cos^(402) α)(sinα)^(1−(1/(50))) ×cosαdα)/(∫_0 ^(π/2) cos^(404) ×sin^(1−(1/(50))) ×cosαdα)) ((∫_0 ^(π/2) (sinα)^(1−(1/(50))) ×(cosα)^(403) dα)/(∫_0 ^(π/2) (sinα)^(1−(1/(50))) ×(cosα)^(405) dα)) now formulA 2∫_0 ^(π/2) sin^(2p−1) αcos^(2q−1) αdα=((⌈(p)⌈(q))/(⌈(p+q))) 2p−1=1−(1/(50)) →2p=2−(1/(50)) →p=1−(1/(100)) 2q−1=403→2q=404→q=202 2q_1 −1=405→2q_1 =406→q_1 =203 q_1 =q+1→⌈(q_1 )=⌈(q+1)→q⌈(q) p+q_1 =p+q+1→⌈(p+q_1 )=⌈(p+q+1)→(p+q)⌈(p+q) ((2∫^(π/2) sin^(2p−1) αcos^(2q−1) αdα)/(2∫_0 ^(π/2) sin^(2p−1) αcos^(2q_1 −1) αdα)) =(((⌈(p)⌈(q))/(⌈(p+q)))/((⌈(p)⌈(q_1 ))/(⌈(p+q_1 )))) =((⌈(p)⌈(q)⌈(p+q_1 ))/(⌈(p)⌈(q_1 )⌈(p+q))) =((⌈(q)×(p+q)⌈(p+q))/(q⌈(q)×⌈(p+q))) =((p+q)/q)=((1−(1/(100))+202)/(202))=((100−1+20200)/(20200))=((20300−1)/(20200)) =((20299)/(20200)) Rahul pls check...$
	$1) \frac{\int_{0}^{1} {1 - {(1 - x^{2})}^{a}}^{2 a + 1} . x d x}{\int_{0}^{1} {1 - {(1 - x^{2})}^{a}}^{2 a + 2} . x d x} a = 100$ $t = 1 - x^{2} d t = - 2 x d x$ $\frac{\frac{1}{2} \int_{1}^{0} {(1 - t^{a})}^{2 a + 1} \times - d t}{\frac{1}{2} \int_{1}^{0} {(1 - t^{a})}^{2 a + 2} \times - d t}$ $\frac{\int_{0}^{1} {(1 - t^{a})}^{2 a + 1} d t}{\int_{0}^{1} {(1 - t^{a})}^{2 a + 2} d t}$ $\frac{\int_{0}^{1} {(1 - t^{100})}^{201} d t}{\int_{0}^{1} {(1 - t^{100})}^{202} d t}$ $l o o k s a m e t y p e q u e s t i o n o f q u e s t i o n n o 2$ $b u t s o l v i n g o t h e r w a y$ $l e t t^{100} = {s i n}^{2} α \to 100 t^{99} d t = 2 s i n α c o s α d α$ $d t = \frac{2 s i n α c o s α d α}{100 {({s i n}^{2} α)}^{\frac{1}{100}}} = \frac{1}{50} \times {(s i n α)}^{1 - \frac{1}{50}} \times c o s α d α$ $\frac{\int_{0}^{\frac{π}{2}} {(1 - {s i n}^{2} α)}^{201} \times \frac{1}{50} \times {(s i n α)}^{1 - \frac{1}{50}} \times c o s α d α}{\int_{0}^{\frac{π}{2}} {(1 - {s i n}^{2} α)}^{202} \times \frac{1}{50} \times {(s i n α)}^{1 - \frac{1}{50}} \times c o s α d α} \frac{}{}$ $\frac{\int^{\frac{π}{2}} {c o s}^{402} α) {(s i n α)}^{1 - \frac{1}{50}} \times c o s α d α}{\int_{0}^{\frac{π}{2}} {c o s}^{404} \times {s i n}^{1 - \frac{1}{50}} \times c o s α d α}$ $\frac{\int_{0}^{\frac{π}{2}} {(s i n α)}^{1 - \frac{1}{50}} \times {(c o s α)}^{403} d α}{\int_{0}^{\frac{π}{2}} {(s i n α)}^{1 - \frac{1}{50}} \times {(c o s α)}^{405} d α}$ $n o w f o r m u l A$ $2 \int_{0}^{\frac{π}{2}} {s i n}^{2 p - 1} α {c o s}^{2 q - 1} α d α = \frac{⌈ (p) ⌈ (q)}{⌈ (p + q)}$ $2 p - 1 = 1 - \frac{1}{50} \to 2 p = 2 - \frac{1}{50} \to p = 1 - \frac{1}{100}$ $2 q - 1 = 403 \to 2 q = 404 \to q = 202$ $2 q_{1} - 1 = 405 \to 2 q_{1} = 406 \to q_{1} = 203$ $q_{1} = q + 1 \to ⌈ (q_{1}) = ⌈ (q + 1) \to q ⌈ (q)$ $p + q_{1} = p + q + 1 \to ⌈ (p + q_{1}) = ⌈ (p + q + 1) \to (p + q) ⌈ (p + q)$ $\frac{2 \int^{\frac{π}{2}} {s i n}^{2 p - 1} α {c o s}^{2 q - 1} α d α}{2 \int_{0}^{\frac{π}{2}} {s i n}^{2 p - 1} α {c o s}^{2 q_{1} - 1} α d α} = \frac{\frac{⌈ (p) ⌈ (q)}{⌈ (p + q)}}{\frac{⌈ (p) ⌈ (q_{1})}{⌈ (p + q_{1})}}$ $= \frac{⌈ (p) ⌈ (q) ⌈ (p + q_{1})}{⌈ (p) ⌈ (q_{1}) ⌈ (p + q)}$ $= \frac{⌈ (q) \times (p + q) ⌈ (p + q)}{q ⌈ (q) \times ⌈ (p + q)}$ $= \frac{p + q}{q} = \frac{1 - \frac{1}{100} + 202}{202} = \frac{100 - 1 + 20200}{20200} = \frac{20300 - 1}{20200}$ $= \frac{20299}{20200} R a h u l p l s c h e c k . . .$
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