evaluate ∫_0 ^( π) e^(sin^2 (u)) du... i use Feynman′s trick to solve integral ∫_0 ^( π) e^(sin^2 (u)) du=I I(t)=∫_0 ^( π) e^(tsin^2 (u)) du I^((1)) (t)=∫_0 ^( π) sin^2 (u)e^(tsin^2 (u)) du ∫_0 ^( π) (1−cos^2 (u))e^(t∙sin^2 (u)) du ∫_0 ^( π) e^(t∙sin^2 (u)) du−∫_0 ^( π) cos^2 (u)e^(t∙sin^2 (u)) du ∫_0 ^( π) e^(t∙sin^2 (u)) du−∫_0 ^( π) sin^2 (u+(π/2))e^(t∙sin^2 (u)) du u+(π/2) =w → du=dw ∫_(π/2) ^((3π)/2) sin^2 (w)e^(t∙cos^2 (w)) dw= ∫_(π/2) ^( ((3π)/2)) sin^2 (w)e^(t(1−sin^2 (w))) dw ∫_(π/2) ^((3π)/2) sin^2 (w)e^t e^(−tsin^2 (w)) dw=e^t ∫_(π/2) ^((3π)/2) sin^2 (w)e^(−tsin^2 (w)) dw I^((1)) (t)=∫_0 ^( π) e^(t∙sin^2 (u)) du−e^t ∫_0 ^( π) sin^2 (u)e^(−t∙sin^2 (u)) du I can′t ...anymore...help


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Question Number 214051 by issac last updated on 25/Nov/24

$evaluate \int_{0}^{π} e^{\sin^{2} (u)} d u . . .$ $i use {Feynman}^{'} s trick to solve integral$ $\int_{0}^{π} e^{\sin^{2} (u)} d u = I$ $I (t) = \int_{0}^{π} e^{t \sin^{2} (u)} d u$ $I^{(1)} (t) = \int_{0}^{π} \sin^{2} (u) e^{t \sin^{2} (u)} d u$ $\int_{0}^{π} (1 - \cos^{2} (u)) e^{t \cdot \sin^{2} (u)} d u$ $\int_{0}^{π} e^{t \cdot \sin^{2} (u)} d u - \int_{0}^{π} \cos^{2} (u) e^{t \cdot \sin^{2} (u)} d u$ $\int_{0}^{π} e^{t \cdot \sin^{2} (u)} d u - \int_{0}^{π} \sin^{2} (u + \frac{π}{2}) e^{t \cdot \sin^{2} (u)} d u$ $u + \frac{π}{2} = w \to d u = d w$ $\int_{\frac{π}{2}}^{\frac{3 π}{2}} \sin^{2} (w) e^{t \cdot \cos^{2} (w)} d w =$ $\int_{\frac{π}{2}}^{\frac{3 π}{2}} \sin^{2} (w) e^{t (1 - \sin^{2} (w))} d w$ $\int_{\frac{π}{2}}^{\frac{3 π}{2}} \sin^{2} (w) e^{t} e^{- t \sin^{2} (w)} d w = e^{t} \int_{\frac{π}{2}}^{\frac{3 π}{2}} \sin^{2} (w) e^{- t \sin^{2} (w)} d w$ $I^{(1)} (t) = \int_{0}^{π} e^{t \cdot \sin^{2} (u)} d u - e^{t} \int_{0}^{π} \sin^{2} (u) e^{- t \cdot \sin^{2} (u)} d u$ $I {can}^{'} t . . . anymore . . . help$
Commented by Ghisom last updated on 26/Nov/24

$Feynman {can}^{'} t help here$
	Answered by Ghisom last updated on 26/Nov/24

	$\underset{0}{\int^{π}} e^{\sin^{2} x} d x = 2 \underset{0}{\int^{π / 2}} e^{\sin^{2} x} d x =$ $[t = 2 x + π]$ $= \sqrt{e} \underset{π}{\int^{2 π}} e^{\frac{1}{2} \cos t} d t = \sqrt{e} \underset{0}{\int^{π}} e^{\frac{1}{2} \cos t} d t$ $Modified Bessel Function of the 1^{st} Kind$ $I_{n} (z) = \frac{1}{π} \underset{0}{\int^{π}} e^{z \cos θ} \cos (n θ) d θ$ $in our case n = 0 \land z = \frac{1}{2} \Rightarrow$ $\sqrt{e} \underset{0}{\int^{π}} e^{\frac{1}{2} \cos t} d t = \sqrt{e} π I_{0} (\frac{1}{2})$
	Commented by issac last updated on 27/Nov/24

	$w o w t h x$
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