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Question Number 5882 by sanusihammed last updated on 03/Jun/16

\int {c o s h}^{n} (x) d x

p l e a s e h e l p .

Answered by Yozzii last updated on 03/Jun/16

{c o s h}^{n} (x) = {(\frac{e^{x} + e^{- x}}{2})}^{n} n \in N + {0} (a s s u m e d)

{c o s h}^{n} (x) = \frac{1}{2^{n}} \underset{k = 0}{\sum^{n}} {(\begin{matrix} n \\ k \end{matrix}) {(e^{x})}^{n - k} {(e^{- x})}^{k}} (B i n o m i a l t h e o r e m)

{c o s h}^{n} x = 2^{- n} \underset{k = 0}{\sum^{n}} {(\begin{matrix} n \\ k \end{matrix}) e^{x (n - 2 k)}}

F o r t e r m i n d e p e n d e n t o f x, \Rightarrow n - 2 k = 0

o r k = \frac{n}{2} w h e r e k \in Z . S o, i f n i s e v e n

t h e n a t e r m i n d e p e n d e n t o f x e x i s t s

a n d i s g i v e n b y 2^{- n} (\begin{matrix} n \\ n / 2 \end{matrix}) . D e f i n e

t h e s e t R = {n, k \in Z^{⩾}, n \equiv 0 (m o d 2) ∣ 0 ⩽ k ⩽ n a n d k \neq \frac{n}{2}} .

\Rightarrow \int {c o s h}^{n} x d x = \int {2^{- n} \sum_{k \in R} (\begin{matrix} n \\ k \end{matrix}) e^{x (n - 2 k)} + 2^{- n} (\begin{matrix} n \\ n / 2 \end{matrix})} d x

\int {c o s h}^{n} x d x = 2^{- n} {\sum_{k \in R} [(\begin{matrix} n \\ k \end{matrix}) \frac{e^{x (n - 2 k)}}{n - 2 k}] + x (\begin{matrix} n \\ n / 2 \end{matrix})} + C

T h i s i s v a l i d f o r n o n - z e r o e v e n n .

I f n = 0 \Rightarrow \int {c o s h}^{n} x d x = x + C

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I f n i s o d d t h e n k = \frac{n}{2} \notin Z, s o t h a t

n o t e r m i n d e p e n d e n t o f x e x i s t s .

∴ \int {c o s h}^{n} x d x = 2^{- n} \underset{k = 0}{\sum^{n}} [(\begin{matrix} n \\ k \end{matrix}) \frac{e^{x (n - 2 k)}}{n - 2 k}] + C

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I n a l l, f o r n \in N + {0}, w h e r e R = {k, n \in Z^{⩾}, n \equiv 0 (m o d 2) ∣ 0 ⩽ k ⩽ n a n d k \neq \frac{n}{2}}

\int {c o s h}^{n} (x) d x = {\begin{cases} 2^{- n} {\sum_{k \in R} [(\begin{array}{c} n \\ k \end{array}) \frac{e^{x (n - 2 k)}}{n - 2 k}] + (\begin{array}{c} n \\ n / 2 \end{array}) x} + C i f n \equiv 0 (m o d 2) a n d n ⩾ 2 \\ 2^{- n} \underset{k = 0}{\sum^{n}} [(\begin{array}{c} n \\ k \end{array}) \frac{e^{x (n - 2 k)}}{n - 2 k}] + C i f n i s o d d \\ x + C i f n = 0 \end{cases}

C i s a c o n s t a n t o f i n t e g r a t i o n .

Commented by Yozzii last updated on 03/Jun/16

S u p p o s e n \in Z^{-} \Rightarrow l e t n = - r w h e r e r \in Z^{+} .

\Rightarrow I (n) = I (- r) = \int {c o s h}^{- r} x d x

I (- r) = \int {s e c h}^{r} x d x = 2^{r} \int \frac{d x}{{(e^{x} + e^{- x})}^{r}}

I (- r) = 2^{r} \int \frac{e^{2 r x}}{{(1 + e^{2 x})}^{r}} d x = 2^{r} \int {(\frac{e^{x}}{1 + e^{2 x}})}^{r} d x

u = e^{x} \Rightarrow d u = e^{x} d x \Rightarrow d x = \frac{1}{u} d u

\Rightarrow I (- r) = 2^{r} \int {(\frac{u}{1 + u^{2}})}^{r} \frac{1}{u} d u

I (- r) = 2^{r} \int \frac{u^{r - 1}}{{(1 + u^{2})}^{r}} d u

\dots

Commented by sanusihammed last updated on 03/Jun/16

T h a n k s s i r .

Commented by sanusihammed last updated on 03/Jun/16

Commented by sanusihammed last updated on 03/Jun/16

P l e a s e c h e c k t h a t r e s u l t . h o w c o m e ?

Commented by Yozzii last updated on 03/Jun/16

T h e f u n c t i o n F (\dots) i s a h y p e r g e o m e t r i c

f u n c t i o n . M y l e v e l o f k n o w l e d g e a n d

u m d e r s t a n d i n g o f M a t h e m a t i c s i s

i n s u f f i c i e n t t o p r o v e t h a t r e s u l t .

S o r r y \dots

Commented by sanusihammed last updated on 03/Jun/16

T h a n k s i r e a l l y a p p r e i a t e