0-xsin-4x-9-4x-2-dx-

Question Number 146621 by 777316 last updated on 14/Jul/21

\int_{0}^{\infty} \frac{x s i n (4 x)}{9 + 4 x^{2}} d x =

Answered by mathmax by abdo last updated on 14/Jul/21

Ψ = \int_{0}^{\infty} \frac{xsin (4 x)}{{4 x}^{2} + 9} \Rightarrow Ψ =_{2 x = 3 z} \int_{0}^{\infty} \frac{3 zsin (4 \frac{3 z}{2})}{2.9 (1 + z^{2})} \frac{3}{2} dz

= \frac{1}{4} \int_{0}^{\infty} \frac{zsin (6 z)}{z^{2} + 1} dz = \frac{1}{8} \int_{- \infty}^{+ \infty} \frac{zsin (6 z)}{z^{2} + 1} dz = \frac{1}{8} Im (\int_{- \infty}^{+ \infty} \frac{{ze}^{6 iz}}{z^{2}) 1} dz)

let φ (z) = \frac{{ze}^{6 iz}}{z^{2} + 1} \Rightarrow φ (z) = \frac{{ze}^{6 iz}}{(z - i) (z + i)}

residus th . \Rightarrow \int_{R} φ (z) dz = 2 i π Res (φ, i)

Res (φ, i) = \frac{{ie}^{- 6}}{2 i} \Rightarrow \int_{R} φ (z) dz = 2 i π . \frac{{ie}^{- 6}}{2 i} = \frac{i π}{e^{6}} \Rightarrow

Ψ = \frac{1}{8} \times \frac{π}{e^{6}} = \frac{π}{{8 e}^{6}}

Answered by KINMATICS last updated on 14/Jul/21

\int \frac{xsin 4 x}{9 + {4 x}^{2}} dx = \int \sin 4 x . \frac{x}{{(2 x)}^{2} - {(3 i)}^{2}} dx

\int \frac{xsin 4 x}{9 + {4 x}^{2}} dx = \int \sin 4 x . \frac{1}{4} (\frac{1}{(2 x + 3 i)} + \frac{1}{2 x - 3 i}) dx

\int \frac{xsin 4 x}{9 + {4 x}^{2}} dx = \frac{1}{2} \int \frac{\sin 4 x}{(4 x + 6 i)} dx + \frac{1}{2} \int \frac{\sin 4 x}{4 x - 6 i} dx

For \int \frac{\sin 4 x}{4 x + 6 i} dx, let 4 x + 6 i = u, 4 x = u - 6 i, dx = \frac{du}{4}

for \int \frac{\sin 4 x}{4 x - 6 i} dx, let 4 x - 6 i = y, 4 x = y + 6 i, dx = \frac{dy}{4}

\Rightarrow \int \frac{\sin 4 xdx}{{4 x}^{2} + 9} = \frac{1}{8} \int \frac{\sin (u - 6 i)}{u} du + \frac{1}{8} \int \frac{\sin (y + 6 i)}{y} dy

I = \frac{1}{8} \int \frac{\sin (u) \cos (6 i) - \cos (u) \sin (6 i)}{u} du +

\frac{1}{8} \int \frac{\sin (y) \cos (6 i) + \cos (y) \sin (6 i)}{y} dy

I = \frac{\cos (6 i)}{8} \int \frac{\sin (u)}{u} du - \frac{\sin (6 i)}{8} \int \frac{\cos (u)}{u} du +

\frac{\cos (6 i)}{8} \int \frac{\sin (y)}{y} dy + \frac{\sin (6 i)}{8} \int \frac{\cos (y)}{y} dy

I = \frac{\cos (6 i)}{8} \boldsymbol S i (u) - \frac{\sin (6 i)}{8} Ci (u) +

\frac{\cos (6 i)}{8} Si (y) + \frac{\sin (6 i)}{8} Ci (y) + C

I = \frac{\cos (6 i)}{8} Si (4 x + 6 i) - \frac{\sin (6 i)}{8} Ci (4 x + 6 i)

\frac{\cos (6 i)}{8} Si (4 x - 6 i) + \frac{\sin (6 i)}{8} Ci (4 x - 6 i) + C

By : Kin Tom

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