∫e^(cos(x)) sin(sin(x)) dx


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Question Number 62122 by aliesam last updated on 15/Jun/19

$\int e^{c o s (x)} s i n (s i n (x)) d x$
Commented by MJS last updated on 15/Jun/19

$\int e^{\cos x} \sin \sin x d x = - \frac{i}{2} \int (e^{\frac{e^{i x} - e^{- i x}}{2}} - e^{- \frac{e^{i x} - e^{- i x}}{2}}) e^{\frac{e^{i x} + e^{- i x}}{2}} d x =$ $= - \frac{i}{2} \int (e^{e^{i x}} - e^{e^{- i x}}) d x = - \frac{i}{2} \int e^{e^{i x}} d x + \frac{i}{2} \int e^{e^{- i x}} d x$ $- \frac{i}{2} \int e^{e^{i x}} d x =$ $[t = e^{i x} \to d x = - {ie}^{- i x} d t]$ $= - \frac{1}{2} \int \frac{e^{t}}{t} d t$ $\frac{i}{2} \int e^{e^{- i x}} d x =$ $[t = e^{- i x} \to d x = {ie}^{i x} d t]$ $= - \frac{1}{2} \int \frac{e^{t}}{t} d t$ $\Rightarrow both lead to the incomplete gamma function$
Commented by aliesam last updated on 15/Jun/19

Commented by maxmathsup by imad last updated on 15/Jun/19
$∫ e^(cosx) sin(sinx)dx =Im (∫ e^(cosx) e^(isinx) dx) ∫ e^(cosx +isinx) dx =∫ e^e^(ix) dx changement e^(ix) =z give ix =ln(z) ⇒ x=(1/i)ln(z) ⇒dx =(dz/(iz)) ⇒ I =∫ e^z (dz/(iz)) =−i ∫ (e^z /z) dz e^z =Σ_(n=0) ^∞ (z^n /(n!)) =1+Σ_(n=1) ^∞ (z^n /(n!)) ⇒(e^z /z) =(1/z) +Σ_(n=1) ^∞ (z^(n−1) /(n!)) ⇒ ∫ (e^z /z)dz = ∫ (dz/z) + Σ_(n=) ^∞ ((z^n )/(nn!)) +c =ln(z) +Σ_(n=1) ^∞ (1/(nn!)){cos(nx)+isin(nx)} +c ⇒−i ∫ (e^z /z)dz =−iln(z)−i Σ_(n=1) ^∞ (1/(nn!)){ cos(nx)+isin(nx)} +c =−i(ix) −i Σ_(n=1) ^∞ ((cos(nx))/(nn!)) +Σ_(n=1) ^∞ ((sin(nx))/(nn!)) +c ⇒ ∫ e^(cosx) sin(sinx)dx = Σ_(n=1) ^∞ ((cos(nx))/(nn!))$
$\int e^{c o s x} s i n (s i n x) d x = I m (\int e^{c o s x} e^{i s i n x} d x)$ $\int e^{c o s x + i s i n x} d x = \int e^{e^{i x}} d x c h a n g e m e n t e^{i x} = z g i v e i x = l n (z) \Rightarrow$ $x = \frac{1}{i} l n (z) \Rightarrow d x = \frac{d z}{i z} \Rightarrow I = \int e^{z} \frac{d z}{i z} = - i \int \frac{e^{z}}{z} d z$ $e^{z} = \sum_{n = 0}^{\infty} \frac{z^{n}}{n!} = 1 + \sum_{n = 1}^{\infty} \frac{z^{n}}{n!} \Rightarrow \frac{e^{z}}{z} = \frac{1}{z} + \sum_{n = 1}^{\infty} \frac{z^{n - 1}}{n!} \Rightarrow$ $\int \frac{e^{z}}{z} d z = \int \frac{d z}{z} + \sum_{n =}^{\infty} \frac{z^{n}}{n n!} + c = l n (z) + \sum_{n = 1}^{\infty} \frac{1}{n n!} {c o s (n x) + i s i n (n x)} + c$ $\Rightarrow - i \int \frac{e^{z}}{z} d z = - i l n (z) - i \sum_{n = 1}^{\infty} \frac{1}{n n!} {c o s (n x) + i s i n (n x)} + c$ $= - i (i x) - i \sum_{n = 1}^{\infty} \frac{c o s (n x)}{n n!} + \sum_{n = 1}^{\infty} \frac{s i n (n x)}{n n!} + c \Rightarrow$ $\int e^{c o s x} s i n (s i n x) d x = \sum_{n = 1}^{\infty} \frac{c o s (n x)}{n n!}$
Commented by maxmathsup by imad last updated on 16/Jun/19

$\int e^{c o s x} s i n (s i n x) = \sum_{n = 1}^{\infty} \frac{c o s (n x)}{n n!} + c$
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