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Question Number 158590 by mnjuly1970 last updated on 06/Nov/21

p r o v e t h a t

1 . I = \int_{0}^{\frac{π}{2}} \frac{s i n (x + t a n (x))}{s i n (x)} d x = \frac{π}{2}

2 . J = \int_{0}^{\frac{π}{2}} \frac{s i n (x - t a n (x))}{s i n (x)} d x = (\frac{1}{e} - \frac{1}{2}) π

Answered by qaz last updated on 06/Nov/21

I = \int_{0}^{π / 2} \frac{\sin x \cdot \cos (\tan x) + \cos x \cdot \sin (\tan x)}{\sin x} dx

= \int_{0}^{π / 2} \cos (\tan x) + \frac{\sin (\tan x)}{\tan x} dx \dots \dots \dots \dots \tan x \to x

= \int_{0}^{\infty} (\cos x + \frac{\sin x}{x}) \frac{dx}{1 + x^{2}}

= \int_{0}^{\infty} \frac{\cos x}{1 + x^{2}} + (\frac{1}{x} - \frac{x}{1 + x^{2}}) \sin xdx

= \int_{0}^{\infty} L {\cos x} \cdot L^{- 1} {\frac{1}{1 + x^{2}}} + (\frac{1}{x} - \frac{x}{1 + x^{2}}) \sin xdx

= \int_{0}^{\infty} \frac{x}{1 + x^{2}} \cdot \sin x + (\frac{1}{x} - \frac{x}{1 + x^{2}}) \cdot \sin xdx

= \frac{π}{2}

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J = \int_{0}^{π / 2} \frac{\sin (x - \tan x)}{\sin x} dx

= \int_{0}^{π / 2} \frac{\sin x \cdot \cos (\tan x) - \cos x \cdot \sin (\tan x)}{\sin x} dx

= \int_{0}^{π / 2} \cos (\tan x) - \frac{\sin (\tan x)}{\tan x} dx \dots \dots \dots \tan x \to x

= \int_{0}^{\infty} (\cos x - \frac{\sin x}{x}) \frac{dx}{1 + x^{2}}

= \int_{0}^{\infty} (\frac{2 \cos x}{1 + x^{2}} - \frac{\sin x}{x}) dx

= \frac{π}{e} - \frac{π}{2}

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\int_{0}^{\infty} \frac{\sin x}{x} dx = \int_{0}^{\infty} L {\sin x} \cdot L^{- 1} {\frac{1}{x}} dx

= \int_{0}^{\infty} \frac{1}{1 + x^{2}} \cdot 1 dx = \frac{π}{2}

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\int_{0}^{\infty} \frac{L {\cos tx}}{1 + x^{2}} dx = \int_{0}^{\infty} \frac{s}{(s^{2} + x^{2}) (1 + x^{2})} dx = \frac{π}{2 (1 + s)}

\Rightarrow \int_{0}^{\infty} \frac{\cos x}{1 + x^{2}} dx = L^{- 1} {\frac{π}{2 (1 + s)}}_{t = 1} = \frac{π}{2} e^{- 1}

Commented by puissant last updated on 06/Nov/21

Ω = \int_{0}^{\infty} \frac{c o s x}{x^{2} + 1} d x = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{c o s x}{x^{2} + 1} d x

= \frac{1}{2} R e (\int_{- \infty}^{+ \infty} \frac{e^{i x}}{x^{2} + 1} d x) = \frac{1}{2} (2 i π ∙ r e s (\frac{e^{i x}}{x^{2} + 1}, + i))

Ω = i π ∙ \lim_{x \to + i} (\frac{e^{i π}}{x + i}) = i π \times \frac{e^{- 1}}{2 i} = \frac{π}{2 e} . .

Ω = \int_{0}^{\infty} \frac{c o s x}{x^{2} + 1} d x = \frac{π}{2 e} . .

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