Integration durch Substitution

a)

(4x) * (1+2x^2)^-0,5 dx Grenzen 2 und 0 durch Substitution.

Integral

$\displaystyle \int \limits_0^{2}\dfrac{4x}{\sqrt{1+2x^2}}\ dx = \ ?$

$\begin{array}{|lrcl|} \hline \displaystyle \int \limits_0^{2}\dfrac{4x}{\sqrt{1+2x^2}}\ dx =\ ? \\\\ & \text{Substitution:}\\ & \mathbf{z = \sqrt{1+2x^2}} &=& (1+2x^2)^{\frac12} \\ & dz &=& \frac12 \cdot (1+2x^2)^{\frac12 - 1} \cdot 2\cdot 2x \ dx \\ & dz &=& 2x \cdot (1+2x^2)^{-\frac12} \ dx \\ & dz &=& \dfrac{ 2x } { \sqrt{1+2x^2} } \ dx \\ & \mathbf{dx} &\mathbf{=}& \mathbf{\dfrac{ \sqrt{1+2x^2} } { 2x } \ dz} \\\\ & \text{Neue Grenzen:}\\ & x&=&0 \quad \Rightarrow \quad z = \sqrt{1+2\cdot 0^2} = 1 \\ & x&=&2 \quad \Rightarrow \quad z = \sqrt{1+2\cdot 2^2} = 3 \\ \hline \end{array}$

$\begin{array}{|rcl|} \hline \displaystyle \mathbf{\int \limits_0^{2} \dfrac{4x}{\sqrt{1+2x^2}}\ dx } \\\\ &=& \displaystyle \int \limits_{z=1}^{z=3}\dfrac{4x}{\sqrt{1+2x^2}}\cdot \dfrac{ \sqrt{1+2x^2} } { 2x } \ dz \\\\ &=& \displaystyle \int \limits_{z=1}^{z=3} 2 \ dz \\\\ &=& \displaystyle 2\int \limits_{z=1}^{z=3} \ dz \\\\ &=& \displaystyle 2[z]_{z=1}^{z=3} \\\\ &=& \displaystyle 2[3-1] \\\\ &=& \displaystyle 2\cdot 2 \\\\ \mathbf{\displaystyle \int \limits_0^{2}\dfrac{4x}{\sqrt{1+2x^2}}\ dx }& \mathbf{=} & \mathbf{4} \\ \hline \end{array}$

c)
x*sin(x^2)dx    Grenzen: 1 und 0 durch Substitution.

Integral

$\displaystyle \int \limits_{0}^{1} x*\sin(x^2) \ dx = \ ?$

$\begin{array}{|lrcl|} \hline \displaystyle \int \limits_{0}^{1} x*\sin(x^2) \ dx = \ ? \\ & \text{Substitution:}\\ & \mathbf{z} &\mathbf{=}& \mathbf{x^2 } \\ & dz &=& 2x \ dx \\ & \mathbf{dx} &\mathbf{=}& \mathbf{\dfrac{1} { 2x } \ dz} \\ \hline \end{array}$

Stammfunktion:

$\begin{array}{|lcl|} \hline \displaystyle \int x*\sin(x^2) \ dx \\\\ &=& \displaystyle \int x*\sin(z) \dfrac{1} { 2x } \ dz \\\\ &=& \displaystyle \dfrac12 \cdot \int \sin(z) \ dz \\\\ &=& \displaystyle \dfrac12 \cdot \left(- \cos(z) \right) +c \\\\ & \text{Rücksubstitution:} \\ &=& \displaystyle \dfrac12 \cdot \left(- \cos(x^2) \right) +c \\\\ &=& \displaystyle -\dfrac12 \cdot \left( \cos(x^2) \right) +c \\\\ \hline \mathbf{\displaystyle \int \limits_{-1}^{1} x*\sin(x^2) \ dx} \\ &=& \displaystyle -\dfrac12 \cdot \left[ \cos(x^2) \right]_{0}^{1} \quad |\quad x \text{ in radiant} \\\\ &=& \displaystyle -\dfrac12 \cdot \left[ \cos(1^2)-\cos(0^2) \right] \\\\ &=& \displaystyle -\dfrac12 \cdot \left( \cos(1)-\cos(0) \right) \\\\ &=& \displaystyle -\dfrac12 \cdot \left( 0.54030230587-1 \right) \\\\ &=& \displaystyle -\dfrac12 \cdot \left( -0.45969769413 \right) \\\\ &=& \displaystyle \dfrac12 \cdot \left( 0.45969769413 \right) \\\\ \mathbf{\displaystyle \int \limits_{-1}^{1} x*\sin(x^2) \ dx} &\mathbf{=}& \mathbf{\displaystyle 0.22984884707} \\ \hline \end{array}$

b)

(-2x) * (4-3x^2)^-2 dx Grenzen 1 und -1 durch Substitution.

Integral

$\displaystyle \int \limits_{-1}^{1} \dfrac{-2x}{ (4-3x^2)^2 } \ dx = \ ?$

$\begin{array}{|lrcl|} \hline \displaystyle \int \limits_{-1}^{1} \dfrac{-2x}{ (4-3x^2)^2 } \ dx = \ ? \\ & \text{Substitution:}\\ & \mathbf{z} &\mathbf{=}& \mathbf{ 4-3x^2 } \\ & dz &=& -6x \ dx \\ & \mathbf{dx} &\mathbf{=}& \mathbf{-\dfrac{1} { 6x } \ dz} \\ \hline \end{array}$

Stammfunktion:

$\begin{array}{|lcl|} \hline \displaystyle \int \dfrac{-2x}{ (4-3x^2)^2 } \ dx \\\\ &=& \displaystyle \int \left( \dfrac{-2x}{ z^2 }\right) \cdot \left(-\dfrac{1} { 6x } \right) \ dz \\\\ &=& \displaystyle \dfrac13\cdot \int \dfrac{1}{ z^2 } \ dz \\\\ &=& \displaystyle \dfrac13\cdot \int z^{-2} \ dz \\\\ &=& \displaystyle \dfrac13\cdot\left( \dfrac{z^{-2+1} }{-2+1} \right) +c \\\\ &=& \displaystyle -\dfrac13\cdot z^{-1} +c \\\\ &=& \displaystyle -\dfrac{1}{3z} +c \\\\ & \text{Rücksubstitution:} \\ &=& \displaystyle -\dfrac{1}{3(4-3x^2)} +c \\\\ \hline \mathbf{\displaystyle \int \limits_{-1}^{1} \dfrac{-2x}{ (4-3x^2)^2 } \ dx} \\ &=& \displaystyle -\dfrac13 \cdot \left[ \dfrac{1}{4-3x^2} \right]_{-1}^{1} \\\\ &=& \displaystyle -\dfrac13 \cdot \left[ \dfrac{1}{4-3\cdot 1^2} - \dfrac{1}{4-3\cdot (-1)^2} \right] \\\\ &=& \displaystyle -\dfrac13 \cdot \left[ \dfrac{1}{1} - \dfrac{1}{1} \right] \\\\ &=& \displaystyle -\dfrac13 \cdot \left[ 1 - 1 \right] \\\\ &=& \displaystyle -\dfrac13 \cdot 0 \\\\ &=& \displaystyle 0 \\\\ \mathbf{\displaystyle \int \limits_{-1}^{1} \dfrac{-2x}{ (4-3x^2)^2 } \ dx} &\mathbf{=}& \mathbf{0} \\ \hline \end{array}$