find-0-1-cos-n-arcosx-dx-with-n-integr-natural-

Question Number 49956 by maxmathsup by imad last updated on 12/Dec/18

f i n d \int_{0}^{1} c o s (n a r c o s x) d x w i t h n i n t e g r n a t u r a l .

Commented by Abdo msup. last updated on 13/Dec/18

w e h a v e c o s (n a r c o s x) + i s i n (n a r c o s x)

= {(c o s (a r c o s x) + i s i n (a r c o s x))}^{n}

= {(x + i \sqrt{1 - x^{2}})}^{n} a n d \int_{0}^{1} c o s (n a r c o s x) d x

= R e (\int_{0}^{1} {(x + i \sqrt{1 - x^{2}})}^{n} d x) b u t

{(x + i \sqrt{1 - x^{2}})}^{n} = \sum_{k = 0}^{n} C_{n}^{k} {(i \sqrt{1 - x^{2}})}^{k} x^{n - k}

= \sum_{k = 0}^{n} i^{k} C_{n}^{k} {(1 - x^{2})}^{\frac{k}{2}} x^{n - k}

= \sum_{p = 0}^{[\frac{n}{2}]} {(- 1)}^{p} C_{n}^{2 p} {(1 - x^{2})}^{p} x^{n - 2 p}

+ \sum_{p = 0}^{[\frac{n - 1}{2}]} i {(- 1)}^{p} C_{n}^{2 p + 1} {(1 - x^{2})}^{\frac{2 p + 1}{2}} x^{n - 2 p - 1} \Rightarrow

\int_{0}^{1} c o s (n a r c o s x) d x = (\int_{0}^{1} \sum_{p = 0}^{[\frac{n}{2}]} {(- 1)}^{p} C_{n}^{2 p} {(1 - x^{2})}^{p} x^{n - 2 p} d x)

= \sum_{p = 0}^{[\frac{n}{2}]} {(- 1)}^{p} C_{n}^{2 p} \int_{0}^{1} {(1 - x^{2})}^{p} x^{n - 2 p} d x c h a n g e m e n t

x = s i n t g i v e

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

= \int_{0}^{\frac{π}{2}} {c o s}^{2 p + 1} t {s i n}^{n - 2 p} t d t \dots b e c o n t i n u e d \dots

Answered by Smail last updated on 13/Dec/18

t = a r c o s (x) \Rightarrow d t = \frac{- d x}{\sqrt{1 - x^{2}}} = \frac{- d x}{\sqrt{1 - {c o s}^{2} t}}

d x = - s i n (t) d t

A = - \int_{π / 2}^{0} s i n (t) c o s (n t) d t

= \int_{0}^{π / 2} s i n (t) c o s (n t) d t

b y p a r t s

u = c o s (n t) \Rightarrow u^{'} = - n s i n (n t)

v^{'} = s i n (t) \Rightarrow v = - c o s (t)

A = - {[c o s (t) c o s (n t)]}_{0}^{π / 2} - n \int_{0}^{π / 2} s i n (n t) c o s (t) d t

b y p a r t s

u = s i n (n t) \Rightarrow u^{'} = n c o s (n t)

v^{'} = c o s (t) \Rightarrow v = s i n (t)

A = 1 - n {[s i n (t) s i n (n t)]}_{0}^{π / 2} + n^{2} \int_{0}^{π / 2} s i n (t) c o s (n t) d t

A = 1 - n s i n (\frac{n π}{2}) + n^{2} A

A (1 - n^{2}) = 1 - n s i n (\frac{n π}{2})

A = \frac{n s i n (\frac{n π}{2}) - 1}{n^{2} - 1}

Commented by Abdo msup. last updated on 13/Dec/18

t h a n k s s i r S m a i l . .

Commented by Smail last updated on 14/Dec/18

y o u a r e q u i t e w e l c o m e

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