intregation-collected-problem-

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Question Number 124076 by TANMAY PANACEA last updated on 30/Nov/20

intregation...collected problem

i n t r e g a t i o n \dots c o l l e c t e d p r o b l e m

Commented by TANMAY PANACEA last updated on 30/Nov/20

Commented by TANMAY PANACEA last updated on 30/Nov/20

Commented by TANMAY PANACEA last updated on 30/Nov/20

Commented by TANMAY PANACEA last updated on 30/Nov/20

Commented by TANMAY PANACEA last updated on 30/Nov/20

calculus collected roblem

c a l c u l u s c o l l e c t e d r o b l e m

Commented by TANMAY PANACEA last updated on 30/Nov/20

Commented by TANMAY PANACEA last updated on 30/Nov/20

Commented by TANMAY PANACEA last updated on 30/Nov/20

Commented by TANMAY PANACEA last updated on 30/Nov/20

Commented by TANMAY PANACEA last updated on 30/Nov/20

Commented by TANMAY PANACEA last updated on 30/Nov/20

Commented by TANMAY PANACEA last updated on 30/Nov/20

Commented by Dwaipayan Shikari last updated on 30/Nov/20

∫_0 ^a (dx/((a^n −x^n )^(1/n) ))=(1/a)∫_0 ^a (1/((1−(x^n /a^n ))^(1/n) ))dx u=(x/a) ⇒ =∫_0 ^1 (1/((1−u^n )^(1/n) ))du =(1/n)∫_0 ^1 u^(1−n) (1−t)^(−(1/n)) dt u^n =t⇒nu^(n−1) =(dt/du) =(1/n)∫_0 ^1 t^((1/n)−1) (1−t)^(−(1/n)) =(1/n).((Γ((1/n))Γ(1−(1/n)))/(Γ(1)))=(π/(nsin((π/n))))

\int_{0}^{a} \frac{d x}{{(a^{n} - x^{n})}^{\frac{1}{n}}} = \frac{1}{a} \int_{0}^{a} \frac{1}{{(1 - \frac{x^{n}}{a^{n}})}^{\frac{1}{n}}} d x u = \frac{x}{a} \Rightarrow

= \int_{0}^{1} \frac{1}{{(1 - u^{n})}^{\frac{1}{n}}} d u = \frac{1}{n} \int_{0}^{1} u^{1 - n} {(1 - t)}^{- \frac{1}{n}} d t u^{n} = t \Rightarrow {n u}^{n - 1} = \frac{d t}{d u}

= \frac{1}{n} \int_{0}^{1} t^{\frac{1}{n} - 1} {(1 - t)}^{- \frac{1}{n}} = \frac{1}{n} . \frac{Γ (\frac{1}{n}) Γ (1 - \frac{1}{n})}{Γ (1)} = \frac{π}{n s i n (\frac{π}{n})}

Commented by TANMAY PANACEA last updated on 30/Nov/20

thank you but question no 22 pls

t h a n k y o u b u t q u e s t i o n n o 22 p l s

Commented by mnjuly1970 last updated on 30/Nov/20

note:: Γ(x)Γ(1−x)=(π/(sin(πx))) note:: ∫_0 ^( π) ln(sin(x))dx=−πln(2) Ω=∫_0 ^( 1) ln(Γ(x))dx Ω=∫_0 ^( 1) ln(Γ(1−x))dx 2Ω=∫_0 ^( 1) ln[(Γ(x)Γ(1−x))]dx 2Ω =^(euler reflection formula) ∫_0 ^( 1) ln((π/(sin(πx))))dx =ln(π)−∫_0 ^( 1) ln(sin(πx))dx =^(πx=t) ln(π)−(1/π)∫_0 ^( π) ln(sin(x))dx =ln(π)−(1/π)(−πln(2)) =ln(π)+ln(2)=ln(2π) ∴ Ω=(1/2)ln(2π)=ln((√(2π)) )✓

n o t e :: Γ (x) Γ (1 - x) = \frac{π}{s i n (π x)}

n o t e :: \int_{0}^{π} l n (s i n (x)) d x = - π l n (2)

Ω = \int_{0}^{1} l n (Γ (x)) d x

Ω = \int_{0}^{1} l n (Γ (1 - x)) d x

2 Ω = \int_{0}^{1} l n [(Γ (x) Γ (1 - x))] d x

2 Ω \overset{e u l e r r e f l e c t i o n f o r m u l a}{=} \int_{0}^{1} l n (\frac{π}{s i n (π x)}) d x

= l n (π) - \int_{0}^{1} l n (s i n (π x)) d x

\overset{π x = t}{=} l n (π) - \frac{1}{π} \int_{0}^{π} l n (s i n (x)) d x

= l n (π) - \frac{1}{π} (- π l n (2))

= l n (π) + l n (2) = l n (2 π)

∴ Ω = \frac{1}{2} l n (2 π) = l n (\sqrt{2 π}) ✓

Commented by TANMAY PANACEA last updated on 30/Nov/20

excellent

e x c e l l e n t

Commented by mnjuly1970 last updated on 30/Nov/20

you are welcom sir tanmay..

y o u a r e w e l c o m s i r t a n m a y . .

Commented by Dwaipayan Shikari last updated on 30/Nov/20

Γ(s)Γ(1−s)=(π/(sinπs)) log(Γ(s)+log(Γ(1−s))=log(π)−log(sinπs) ∫_0 ^1 log(Γ(s))=∫_0 ^1 log(Γ(1−s))ds ∫_0 ^1 2log(Γ(s))=(1/2)∫_0 ^1 log(π)ds−(1/2)∫_0 ^1 log(sinπs)ds ∫_0 ^1 log(Γ(s))=(1/2)log(π)−(1/(2π))∫_0 ^π log(sinu)du u=2t⇒1=2(dt/du) =(1/2)log(π)−(2/(2π))∫_0 ^(π/2) log(2)+log(sint)+log(cost) =(1/2)log(π)−(1/2)log(2)+(1/2)log(4) =(1/2)log(2π)

Γ (s) Γ (1 - s) = \frac{π}{s i n π s}

l o g (Γ (s) + l o g (Γ (1 - s)) = l o g (π) - l o g (s i n π s)

\int_{0}^{1} l o g (Γ (s)) = \int_{0}^{1} l o g (Γ (1 - s)) d s

\int_{0}^{1} 2 l o g (Γ (s)) = \frac{1}{2} \int_{0}^{1} l o g (π) d s - \frac{1}{2} \int_{0}^{1} l o g (s i n π s) d s

\int_{0}^{1} l o g (Γ (s)) = \frac{1}{2} l o g (π) - \frac{1}{2 π} \int_{0}^{π} l o g (s i n u) d u u = 2 t \Rightarrow 1 = 2 \frac{d t}{d u}

= \frac{1}{2} l o g (π) - \frac{2}{2 π} \int_{0}^{\frac{π}{2}} l o g (2) + l o g (s i n t) + l o g (c o s t)

= \frac{1}{2} l o g (π) - \frac{1}{2} l o g (2) + \frac{1}{2} l o g (4)

= \frac{1}{2} l o g (2 π)

Commented by Dwaipayan Shikari last updated on 30/Nov/20

∫_0 ^(π/2) (1/( (√(acos^4 x+bsin^4 x))))dx =∫_0 ^(π/2) ((sec^2 x)/( (√(a+btan^4 x))))dx=∫_0 ^∞ (dt/( (√(a+bt^4 )))) a+bt^4 =u^2 ⇒2bt^3 =u(du/dt) =(1/2)∫_(√a) ^∞ ((udu)/(ubt^3 ))=(1/(2b))∫_(√a) ^∞ (du/((((u^2 −a)/b))^(3/4) )) =(1/(2 (b)^(1/4) ))∫_0 ^∞ (du/((u^2 −a)^(3/4) )) u^2 −a=p⇒2u=(dp/du) =(1/(4(b)^(1/4) ))∫_0 ^∞ (dp/(p^(3/4) (a+p)^(1/2) ))=(1/(4((ab))^(1/4) ))∫_0 ^∞ (1/(Φ^(3/4) (1+Φ)^(1/2) ))dΦ Φ=(p/a) =(1/(4((ab))^(1/4) ))∫_0 ^1 Λ^(−(3/4)) (1−Λ)^(−(3/4)) dΛ (Φ/(1+Φ))=Λ =(1/(4((ab))^(1/4) )).Γ^2 ((1/4)).(1/(Γ((1/2))))=((Γ^2 ((1/4)))/(4((π^2 ab))^(1/4) ))

\int_{0}^{\frac{π}{2}} \frac{1}{\sqrt{{a c o s}^{4} x + {b s i n}^{4} x}} d x

= \int_{0}^{\frac{π}{2}} \frac{{s e c}^{2} x}{\sqrt{a + {b t a n}^{4} x}} d x = \int_{0}^{\infty} \frac{d t}{\sqrt{a + {b t}^{4}}} a + {b t}^{4} = u^{2} \Rightarrow 2 {b t}^{3} = u \frac{d u}{d t}

= \frac{1}{2} \int_{\sqrt{a}}^{\infty} \frac{u d u}{{u b t}^{3}} = \frac{1}{2 b} \int_{\sqrt{a}}^{\infty} \frac{d u}{{(\frac{u^{2} - a}{b})}^{\frac{3}{4}}}

= \frac{1}{2 \sqrt[4]{b}} \int_{0}^{\infty} \frac{d u}{{(u^{2} - a)}^{\frac{3}{4}}} u^{2} - a = p \Rightarrow 2 u = \frac{d p}{d u}

= \frac{1}{4 \sqrt[4]{b}} \int_{0}^{\infty} \frac{d p}{p^{\frac{3}{4}} {(a + p)}^{\frac{1}{2}}} = \frac{1}{4 \sqrt[4]{a b}} \int_{0}^{\infty} \frac{1}{Φ^{\frac{3}{4}} {(1 + Φ)}^{\frac{1}{2}}} d Φ Φ = \frac{p}{a}

= \frac{1}{4 \sqrt[4]{a b}} \int_{0}^{1} Λ^{- \frac{3}{4}} {(1 - Λ)}^{- \frac{3}{4}} d Λ \frac{Φ}{1 + Φ} = Λ

= \frac{1}{4 \sqrt[4]{a b}} . Γ^{2} (\frac{1}{4}) . \frac{1}{Γ (\frac{1}{2})} = \frac{Γ^{2} (\frac{1}{4})}{4 \sqrt[4]{π^{2} a b}}

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