Question

нужна с определенными интегралами очень нужно заранее

Answer 1

$1)\ \ \int\limits_0^1\, \dfrac{dx}{(13x+1)^4}=\Big[\ t=13x+1\ ,\ dt=13\, dx\ \Big]=\dfrac{1}{13}\int \limits _1^{14}t^{-4}\, dt=\\\\\\=\dfrac{1}{13}\cdot \dfrac{t^{-3}}{-3}\, \Big|_1^{14}=-\dfrac{1}{39\, \, t^3}\, \Big|_1^{14}=-\dfrac{1}{39}\cdot \Big(\dfrac{1}{14^3}-1\Big)=\dfrac{1}{39}\cdot \dfrac{2743}{2744}=\dfrac{211}{8232}$

$2)\ \ \int\limits^2_0\, \dfrac{4x\, dx}{(x^2-1)^3}=\Big[\ t=x^2-1\ ,\ dt=2x\, dx\ \Big]=2\int\limits^{3}_{-1}\dfrac{dt}{t^3}=2\cdot \dfrac{t^{-2}}{-2}\, \Big|_{-1}^3=\\\\\\=-\dfrac{1}{t^2}\Big|_{-1}^3=-\dfrac{1}{9}+1=\dfrac{8}{9}$

$3)\ \ \int\limits^{\pi /2}_0\, x\, cosx\, dx=\Big[\ u=x\ ,\ du=dx\ ,\ dv=cosx\, dx\ ,\ v=sinx\ \Big]=\\\\\\=x\cdot sinx\Big|_0^{\pi /2}-\int\limits^{\pi /2}_0\, sinx\, dx=\dfrac{\pi}{2}+cosx\, \Big|_0^{\pi /2}=\dfrac{\pi}{2}-1$

Answer 2

1.

$\int\limits^{1} _ {0} \frac{dx}{ {(13x + 1)}^{4} } = \frac{1}{13} \int\limits^{1} _ {0} {(13x + 1)}^{ - 4} d(13x + 4) = \\ = \frac{1}{13} \times \frac{ {(13x + 1)}^{ - 3} }{( - 3)} | ^{1} _ {0} = - \frac{1}{39 {(13x + 1)}^{3} } | ^{1} _ {0} = \\ = - \frac{1}{39} ( \frac{1}{ {(13 + 1)}^{3} } - \frac{1}{1} ) = - \frac{1}{39} ( \frac{1}{2744} - 1) = \\ = - \frac{1}{39} \times ( - \frac{2743}{2744} ) = \frac{211}{8232}$

2.

$\int\limits^{2} _ {0} \frac{4xdx}{ {( {x}^{2} - 1) }^{3} } = \int\limits^{2} _ {0} \frac{2 \times 2xdx}{ {( {x}^{2} - 1) }^{3} } = \\ = 2 \int\limits^{2} _ {0} \frac{d( {x}^{2} - 1) }{ {( {x}^{2} - 1)}^{3} } = 2 \times \frac{ {( {x}^{2} - 1)}^{ - 2} }{( - 2)} | ^{2} _ {0} = \\ = - \frac{1}{ {( {x}^{2} - 1) }^{2} } | ^{2} _ {0} = \\ = - \frac{1}{ {(4 - 1)}^{2} } + \frac{1}{1} = 1 - \frac{1}{9} = \frac{8}{9}$

3.

$\int\limits ^{ \frac{\pi}{2} } _ {0} x \cos(x) dx \\$

по частям:

$U = x \: \: \: \: \: \: dU = dx \\ dV = \cos(x) \: \: \: V = \sin(x)$

$UV - \int\limits \: VdU= \\ = x \sin( x) | ^{ \frac{\pi}{2} } _ {0} - \int\limits ^{ \frac{\pi}{2} } _ {0} \sin(x) dx = \\ =( x \sin(x) + \cos(x) )| ^{ \frac{\pi}{2} } _ {0} = \\ = \frac{\pi}{2} \times \sin( \frac{\pi}{2} ) + \cos( \frac{\pi}{2} ) - 0 - \cos(0) = \\ = \frac{\pi}{2} + 0 - 1 = \frac{\pi}{2} - 1$