calculate-0-e-2x-1-sinx-dx-

Question Number 145044 by mathmax by abdo last updated on 01/Jul/21

calculate \int_{0}^{\infty} e^{- 2 x} \sqrt{1 + sinx} dx

Answered by qaz last updated on 02/Jul/21

\int_{0}^{\infty} e^{- 2 x} \sqrt{1 + \sin x} dx

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

= L (\sin \frac{x}{2} + \cos \frac{x}{2}) (s = 2)

= \frac{\frac{1}{2} + s}{s^{2} + \frac{1}{4}} ∣_{s = 2}

= \frac{10}{17}

Commented by mathmax by abdo last updated on 03/Jul/21

answer not correct sir gaz

Answered by mnjuly1970 last updated on 02/Jul/21

\sqrt{1 + s i n (x)} = ∣ s i n (\frac{x}{2}) + c o s (\frac{x}{2}) ∣

Answered by mathmax by abdo last updated on 02/Jul/21

I = \int_{0}^{\infty} e^{- 2 x} \sqrt{1 + sinx} dx we have 1 + sinx = {(\cos (\frac{x}{2}) + \sin (\frac{x}{2}))}^{2} \Rightarrow

\sqrt{1 + sinx} =∣ \cos (\frac{x}{2}) + \sin (\frac{x}{2}) ∣ \Rightarrow

I = \int_{0}^{\infty} e^{- 2 x} ∣ \cos (\frac{x}{2}) + \sin (\frac{x}{2}) ∣ dx =_{\frac{x}{2} = t} 2 \int_{0}^{\infty} e^{- 4 t} ∣ cost + sint ∣ dt

= 2 \int_{0}^{\infty} e^{- 4 t} ∣ \sqrt{2} \sin (t + \frac{π}{4}) ∣ dt = 2 \sqrt{2} \int_{0}^{\infty} e^{- 4 t} ∣ \sin (t + \frac{π}{4}) ∣ dt

=_{t + \frac{π}{4} = y} 2 \sqrt{2} \int_{\frac{π}{4}}^{+ \infty} e^{- 4 (y - \frac{π}{4})} ∣ siny ∣ dy

= 2 \sqrt{2} e^{π} \int_{\frac{π}{4}}^{+ \infty} e^{- 4 y} ∣ siny ∣ dy

= 2 \sqrt{2} e^{π} (\int_{0}^{\infty} e^{- 4 y} ∣ siny ∣ dy - \int_{0}^{\frac{π}{4}} e^{- 4 y} siny dy) we have

\int_{0}^{\frac{π}{4}} e^{- 4 y} siny dy = Im (\int_{0}^{\frac{π}{4}} e^{- 4 y + iy} dy)

\int_{0}^{\frac{π}{4}} e^{(- 4 + i) y} dy = {[\frac{1}{- 4 + i} e^{(- 4 + i) y}]}_{0}^{\frac{π}{4}}

= - \frac{1}{4 - i} (e^{(- 4 + i) \frac{π}{4}} - 1)

= - \frac{4 + i}{17} (e^{- π} (\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}) - 1)

= - \frac{1}{17} (4 + i) (\frac{e^{- π}}{\sqrt{2}} + i \frac{e^{- π}}{\sqrt{2}} - 1)

= - \frac{1}{17} (\frac{{4 e}^{- π}}{\sqrt{2}} + 4 i \frac{e^{- π}}{\sqrt{2}} - 4 + i \frac{e^{- π}}{\sqrt{2}} - \frac{e^{- π}}{\sqrt{2}} - i) \Rightarrow

\int_{0}^{\frac{π}{4}} e^{- 4 y} sinydy = - \frac{1}{17} (\frac{{4 e}^{- π}}{\sqrt{2}} + \frac{e^{- π}}{\sqrt{2}} - 1) = \frac{1}{17} (1 - \frac{{5 e}^{- π}}{\sqrt{2}})

\int_{0}^{\infty} e^{- 4 y} ∣ siny ∣ dy = \sum_{n = 0}^{\infty} \int_{n π}^{(n + 1) π} e^{- 4 y} ∣ siny ∣ dy

=_{y = n π + z} \sum_{n = 0}^{\infty} \int_{0}^{π} e^{- 4 (n π + z)} sinz dz (sinz ⩾ 0 on [0, π])

= \sum_{n = 0}^{\infty} e^{- 4 n π} \int_{0}^{π} e^{- 4 z} sinzdz we have

\int_{0}^{π} e^{- 4 z} sinz dz = Im (\int_{0}^{π} e^{- 4 z + iz} dz)

\int_{0}^{π} e^{(- 4 + i) z} dz = {[\frac{1}{- 4 + i} e^{(- 4 + i) z}]}_{0}^{π}

= - \frac{1}{4 - i} (e^{(- 4 + i) π} - 1)

= - \frac{4 + i}{17} (- e^{- 4 π} - 1) = \frac{4 + i}{17} (1 + e^{- 4 π}) \Rightarrow

\int_{0}^{π} e^{- 4 z} sinz dz = \frac{1 + e^{- 4 π}}{17} \Rightarrow

\int_{0}^{\infty} e^{- 4 y} ∣ siny ∣ dy = \frac{1 + e^{- 4 π}}{17} \sum_{n = 0}^{\infty} {(e^{- 4 π})}^{n}

= \frac{1 + e^{- 4 π}}{17} \times \frac{1}{1 - e^{- 4 π}} = \frac{1 + e^{4 π}}{17 (1 - e^{- 4 π})} \Rightarrow

I = 2 \sqrt{2} e^{π} (\frac{1 + e^{4 π}}{17 (1 - e^{4 π})} - \frac{1}{17} (1 - \frac{{5 e}^{- π}}{\sqrt{2}})

= \frac{2 \sqrt{2}}{17} (\frac{1 + e^{4 π}}{1 - e^{4 π}} - 1 + \frac{{5 e}^{- π}}{\sqrt{2}})

Commented by mathmax by abdo last updated on 02/Jul/21

I = \frac{2 \sqrt{2}}{17} (\frac{1 + e^{4 π}}{1 - e^{- 4 π}} - 1 + \frac{{5 e}^{- π}}{\sqrt{2}})

Commented by mnjuly1970 last updated on 03/Jul/21

v e r y n i c e s i r m a x \dots

Commented by qaz last updated on 04/Jul/21

nice solution sir

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