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Question Number 42704 by maxmathsup by imad last updated on 01/Sep/18

f (x) = \frac{e^{3 x} + e^{- 3 x}}{2}

1) d e t e r m i n e f^{- 1} (x)

2) c a l c u l a t e \int_{0}^{1} x f (x) d x a n d \int_{0}^{1} f (x) d x

3) c a l c u l a t e \int f^{- 1} (x) d x

4) c a l c u l a t e u_{n} = \int_{0}^{π} f (x) c o s (n x) d x a n d v_{n} = \int_{0}^{n} f (x) s i n (n x) d x

f i n d n a t u r e o f Σ \frac{v_{n}}{u_{n}}

\int_{0}^{1} x f (x) d x = \frac{1}{2} \int_{0}^{1} x e^{3 x} d x + \frac{1}{2} \int_{0}^{1} x e^{- 3 x} d x (b y p a r t s)

= \frac{1}{2} {{[\frac{x}{3} e^{3 x}]}_{0}^{1} - \frac{1}{3} \int_{0}^{1} e^{3 x} d x + {[- \frac{x}{3} e^{- 3 x}]}_{0}^{1} + \frac{1}{3} \int_{0}^{1} e^{- 3 x} d x}

= \frac{1}{2} {\frac{e^{3}}{3} - \frac{1}{9} (e^{3} - 1) - \frac{e^{- 3}}{3} - \frac{1}{9} (e^{- 3} - 1)}

Commented by maxmathsup by imad last updated on 01/Sep/18

1) f (x) = c h (3 x) = y \Leftrightarrow 3 x = a r g c h (y) \Leftrightarrow x = \frac{1}{3} a r g c h (y) = \frac{1}{3} l n (y + \sqrt{y^{2} - 1}) \Rightarrow

f^{- 1} (x) = \frac{1}{3} l n (y + \sqrt{y^{2} - 1})

2) \int_{0}^{1} f (x) d x = \frac{1}{2} \int_{0}^{1} (e^{3 x} + e^{- 3 x}) d x = \frac{1}{2} {[\frac{1}{3} e^{3 x} - \frac{1}{3} e^{- 3 x}]}_{0}^{1}

= \frac{1}{6} {e^{3} - e^{- 3}}

Commented by maxmathsup by imad last updated on 01/Sep/18

3) l e t I = \int f^{- 1} (x) d x l e t f^{- 1} (x) = t \Rightarrow x = f (t) \Rightarrow

I = \int t f^{^{'}} (t) d t = t f (t) - \int f (t) d t a n d

\int f (t) d t = \int \frac{e^{3 t} + e^{- 3 t}}{2} d t = \frac{1}{6} e^{3 t} - \frac{1}{6} e^{- 3 t} + c \Rightarrow

I = t f (t) - \frac{1}{6} e^{3 t} + \frac{1}{6} e^{- 3 t} b u t t = f^{- 1} (x) = \frac{1}{3} l n (x + \sqrt{x^{2} - 1}) \Rightarrow

I = \frac{x}{3} l n (x + \sqrt{x^{2} - 1}) - \frac{1}{6} (x + \sqrt{x^{2} - 1}) + \frac{1}{6 (x + \sqrt{x^{2} - 1})} + c

Commented by maxmathsup by imad last updated on 02/Sep/18

4) w e h a v e u_{n} = \int_{0}^{π} f (x) c o s (n x) d x = \frac{1}{2} \int_{0}^{π} (e^{3 x} + e^{- 3 x}) c o s (n x) d x \Rightarrow

2 u_{n} = R e (\int_{0}^{π} e^{i n x} (e^{3 x} + e^{- 3 x}) d x) = R e (\int_{0}^{π} e^{(3 + i n) x} + e^{(- 3 + i n) x}) d x b u t

\int_{0}^{π} (e^{(3 + i n) x} + e^{(- 3 + i n) x}) d x = {[\frac{1}{3 + i n} e^{(3 + i n) x} + \frac{1}{- 3 + i n} e^{(- 3 + i n) x}]}_{0}^{π}

= \frac{e^{3 π} {(- 1)}^{n}}{3 + i n} + \frac{e^{- 3 π} {(- 1)}^{n}}{- 3 + i n} - \frac{1}{3 + i n} - \frac{1}{- 3 + i n}

= \frac{(3 - i n) e^{3 π} {(- 1)}^{n}}{9 + n^{2}} + \frac{(- 3 - i n) e^{- 3 π} {(- 1)}^{n}}{9 + n^{2}} - \frac{1}{3 + i n} + \frac{1}{3 - i n}

= \frac{{(- 1)}^{n}}{9 + n^{2}} {3 e^{3 π} - i n e^{3 π} - 3 e^{- 3 π} - i n e^{- 3 π}} + \frac{1}{3 - i n} - \frac{1}{3 + i n}

= \frac{3 {(- 1)}^{n}}{9 + n^{2}} {e^{3 π} - e^{- 3 π}} - \frac{n {(- 1)}^{n}}{9 + n^{2}} i {e^{3 π} + e^{- 3 π}} + \frac{2 i n}{9 + n^{2}}

\Rightarrow u_{n} = \frac{3 {(- 1)}^{n} (e^{3 π} - e^{- 3 π})}{9 + n^{2}} a l s o w e h a v e v_{n} = I m (\int_{0}^{π} e^{(3 + i n) x} + e^{(- 3 + i n) x}) d x)

\Rightarrow v_{n} = \frac{2 n - n {(- 1)}^{n}}{9 + n^{2}} .

Commented by maxmathsup by imad last updated on 02/Sep/18

u_{n} = \frac{3 {(- 1)}^{n} (e^{3 π} - e^{- 3 π})}{2 (9 + n^{2})} a n d v_{n} = \frac{2 n - n {(- 1)}^{n} {e^{3 π} + e^{- 3 π}}}{2 (9 + n^{2})}

u_{n} = \frac{3 {(- 1)}^{n} (e^{3 π} - e^{- 3 π})}{2 (9 + n^{2})} a n d v_{n} = \frac{2 n - n {(- 1)}^{n} {e^{3 π} + e^{- 3 π}}}{2 (9 + n^{2})}