sin-2tan-1-x-x-x-the-limit-0-

Question Number 127885 by psyche last updated on 02/Jan/21

\int (\frac{\sin ({2 \tan}^{- 1} (x) + x)}{x}) t h e l i m i t [0, \infty)

Answered by Lordose last updated on 02/Jan/21

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

\sin ({2 \tan}^{- 1} (x) + x) = \sin ({2 \tan}^{- 1} (x)) \cos (x) + \cos ({2 \tan}^{- 1} (x)) \sin (x)

N . B :: (\begin{matrix} \sin ({2 \tan}^{- 1} (x)) = \frac{2 x}{1 + x^{2}} \\ \cos ({2 \tan}^{- 1} (x)) = \frac{1 - x^{2}}{1 + x^{2}} \end{matrix})

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

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

Ω = 2 \cdot \frac{π}{2 e} + \int_{0}^{\infty} \frac{\sin (x)}{x (1 + x^{2})} dx - \int_{0}^{\infty} \frac{\sin (x)}{x} dx + \int_{0}^{\infty} \frac{\sin (x)}{x (1 + x^{2})} dx

Ω = \frac{π}{e} - \frac{π}{2} + 2 Φ

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

Ω = \frac{π}{e} - \frac{π}{2} + π - \frac{π}{e} = \frac{π}{2}

★ \boldsymbol L ϕ rD \boldsymbol \emptyset sE

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