∫_0 ^(2π) xe^(cosx) cos(sinx)dx= 2π^2


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Question Number 98244 by ~blr237~ last updated on 12/Jun/20

$\int_{0}^{2 π} {x e}^{c o s x} c o s (s i n x) d x = 2 π^{2}$
	Answered by maths mind last updated on 15/Jun/20
	$e^(cos(x)+isin(x)) =e^(cos(x)) (cos(sin(x))+isin(sin(x))) ∫_0 ^(2π) xe^(cos(x)+isin(x)) dx =∫_0 ^(2π) xe^e^(ix) dx =∫_0 ^(2π) Σ_(k≥0) x.(e^(ikx) /(k!)) =Σ_(k≥0) (1/(k!))∫_0 ^(2π) xe^(ikx) dx ∫_0 ^(2π) xe^(ikx) dx=[x(e^(ikx) /(ik))]−∫(e^(ik) /(ik))dx =((2π)/(ik)) ,k≠0 if k=0 ∫_0 ^(2π) xdx=2π^2 Re{∫_0 ^(2π) xe^e^(ix) dx}=∫_0 ^(2π) xe^(cos(x)) cos(sin(x))dx =Re[2π^2 +Σ_(k≥1) ((2π)/(ik.k!))] =2π^2 Σ(1/(kk!))=∫(e^x /x)dx=Σ_(k≥0) (x^k /(kk!)) ⇒∫xe^(cos(x)) sin(sin(x))dx=−2πE_i (1)$
	$e^{c o s (x) + i s i n (x)} = e^{c o s (x)} (c o s (s i n (x)) + i s i n (s i n (x)))$ $\int_{0}^{2 π} {x e}^{c o s (x) + i s i n (x)} d x$ $= \int_{0}^{2 π} {x e}^{e^{i x}} d x$ $= \int_{0}^{2 π} \sum_{k ⩾ 0} x . \frac{e^{i k x}}{k!}$ $= \sum_{k ⩾ 0} \frac{1}{k!} \int_{0}^{2 π} {x e}^{i k x} d x$ $\int_{0}^{2 π} {x e}^{i k x} d x = [x \frac{e^{i k x}}{i k}] - \int \frac{e^{i k}}{i k} d x$ $= \frac{2 π}{i k}, k \neq 0$ $i f k = 0$ $\int_{0}^{2 π} x d x = 2 π^{2}$ $R e {\int_{0}^{2 π} {x e}^{e^{i x}} d x} = \int_{0}^{2 π} {x e}^{c o s (x)} c o s (s i n (x)) d x$ $= R e [2 π^{2} + \sum_{k ⩾ 1} \frac{2 π}{i k . k!}]$ $= 2 π^{2}$ $Σ \frac{1}{k k!} = \int \frac{e^{x}}{x} d x = \sum_{k ⩾ 0} \frac{x^{k}}{k k!}$ $\Rightarrow \int {x e}^{c o s (x)} s i n (s i n (x)) d x = - 2 π E_{i} (1)$
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