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Question Number 93907 by abdomathmax last updated on 16/May/20

1) c a l c u l a t e A_{n} = \int_{0}^{\frac{π}{2}} {c o s}^{n} x d x

2) c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} - x + 1)}^{n}}

n i n t e g r n a t u r a l

Commented by mathmax by abdo last updated on 17/May/20

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

= A_{n} - \int_{0}^{\frac{π}{2}} {s i n}^{2} x {c o s}^{n} x d x b u t b y p a r t s u = s i n x a n d v^{^{'}} = s i n x {c o s}^{n} x

\int_{0}^{\frac{π}{2}} {s i n}^{2} x {c o s}^{n} x d x = {[- \frac{1}{n + 1} {c o s}^{n + 1} x s i n x]}_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} \frac{1}{n + 1} {c o s}^{n + 2} x d x \Rightarrow

A_{n + 2} = A_{n} - \frac{1}{n + 1} A_{n + 2} \Rightarrow (1 + \frac{1}{n + 1}) A_{n + 2} = A_{n} \Rightarrow

A_{n + 2} = \frac{n + 1}{n + 2} A_{n} \Rightarrow A_{n} = \frac{n - 1}{n} A_{n - 2}

A_{2 k} = \frac{2 k - 1}{2 k} A_{2 k - 2} \Rightarrow \prod_{k = 1}^{n} \frac{A_{2 k}}{A_{2 k - 2}} = \prod_{k = 1}^{n} \frac{2 k - 1}{2 k} \Rightarrow

\frac{A_{2}}{A_{0}} \times \frac{A_{4}}{A_{2}} \times \dots . \frac{A_{2 n}}{A_{2 n - 2}} = \frac{1.3.5 \dots . . (2 n - 1)}{2^{n} n!} = \frac{1.2.3.4 \dots (2 n - 1) (2 n)}{2^{n} n! 2^{n} n!}

= \frac{(2 n)!}{2^{2 n} {(n!)}^{2}} \Rightarrow A_{2 n} = \frac{(2 n)!}{2^{2 n} {(n!)}^{2}} A_{0} \Rightarrow A_{2 n} = \frac{(2 n)!}{2^{2 n} {(n!)}^{2}} \times \frac{π}{2}

A_{2 n + 1} = \frac{2 n}{2 n + 1} A_{2 n - 1} \Rightarrow \prod_{k = 1}^{n} \frac{A_{2 k + 1}}{A_{2 k - 1}} = \prod_{k = 1}^{n} \frac{2 k}{2 k + 1} \Rightarrow

\frac{A_{3}}{A_{1}} \times \frac{A_{5}}{A_{3}} \times \dots . . \frac{A_{2 n + 1}}{A_{2 n - 1}} = \frac{2^{n} n!}{3.5 \dots . . (2 n + 1)} = \frac{2^{2 n} {(n!)}^{2}}{2.3.4.5 \dots . (2 n) (2 n + 1)}

= \frac{2^{2 n} {(n!)}^{2}}{(2 n + 1)!} \Rightarrow A_{2 n + 1} = \frac{2^{2 n} {(n!)}^{2}}{(2 n + 1)!} \times A_{1} b u t A_{1} = 1 \Rightarrow

A_{2 n + 1} = \frac{2^{2 n} {(n!)}^{2}}{(2 n + 1)!}

Commented by mathmax by abdo last updated on 17/May/20

2) l e t I_{n} = \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} - x + 1)}^{n}} \Rightarrow I_{n} = \int_{- \infty}^{+ \infty} \frac{d x}{{{(x - \frac{1}{2})}^{2} + \frac{3}{4}}^{n}}

=_{x - \frac{1}{2} = \frac{\sqrt{3}}{2} t} {(\frac{4}{3})}^{n} \int_{- \infty}^{+ \infty} \frac{1}{{(t^{2} + 1)}^{n}} \times \frac{\sqrt{3}}{2} d t

= \frac{\sqrt{3}}{2} {(\frac{4}{3})}^{n} \int_{- \infty}^{+ \infty} \frac{d t}{{(t^{2} + 1)}^{n}} = \sqrt{3} {(\frac{4}{3})}^{n} \int_{0}^{\infty} \frac{d t}{{(1 + t^{2})}^{n}}

=_{t = t a n θ} \sqrt{3} {(\frac{4}{3})}^{n} \int_{0}^{\frac{π}{2}} \frac{(1 + {t a n}^{2} θ) d θ}{{(1 + {t a n}^{2} θ)}^{n}}

= \sqrt{3} {(\frac{4}{3})}^{n} \int_{0}^{\frac{π}{2}} \frac{d θ}{{(1 + {t a n}^{2} θ)}^{n - 1}} = \sqrt{3} {(\frac{4}{3})}^{n} \int_{0}^{\frac{π}{2}} {c o s}^{2 n - 2} θ d θ

b u t \int_{0}^{\frac{π}{2}} {c o s}^{2 n} θ d θ = A_{2 n} = \frac{(2 n)!}{2^{2 n} {(n!)}^{2}} \times \frac{π}{2} \Rightarrow

I_{n} = \sqrt{3} {(\frac{4}{3})}^{n} \times \frac{(2 n - 2)!}{2^{2 n - 2} {((n - 1)!)}^{2}} \times \frac{π}{2}

Commented by Ar Brandon last updated on 17/May/20

Amazing .

Please I^{'} ll like to know is A_{2 n} same as

A_{2 n} = \frac{(2 n - 1)!}{(2 n)!} \times \frac{π}{2} ?

Commented by mathmax by abdo last updated on 17/May/20

s i r v e r i f y f o r n = 1, 2, 3 a n d p r o v e i t b y r e c u r r e n c e . . i f i t i s c o r r e c t \dots

Answered by Ar Brandon last updated on 16/May/20

A_{n} = \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \cdot \cdot \frac{n - 1}{n} \cdot \frac{π}{2} if n is even

= \frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7} \cdot \cdot \cdot \frac{n - 1}{n} if n is odd

Answered by Ar Brandon last updated on 16/May/20

2 ∖ B = \int_{- \infty}^{+ \infty} \frac{dx}{{(x^{2} - x + 1)}^{n}} = \int_{- \infty}^{+ \infty} \frac{dx}{{[{(x - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}]}^{n}}

Consider J_{n} = \int_{- \infty}^{+ \infty} \frac{dx}{{(x^{2} + u^{2})}^{n}} \Rightarrow J_{n} = \frac{2 n - 3}{{2 u}^{2} (n - 1)} J_{n - 1}

\Rightarrow u = \frac{\sqrt{3}}{2} \Rightarrow B_{n} = \frac{4 n - 6}{3 (n - 1)} B_{n - 1}