Question

Integrate $\displaystyle {e}^{ax} sin(bx) \: and \: {e}^{bx} cos(ax)$
using Euler's method. Brief on Euler's Method.

Thank you :)​

Answer 1

Given : $\[\int{{{e}^{ax}}\sin \left( bx \right)}dx\]$

To Find : integrate

Solution:

Assume that I is the integral of $\[\int{{{e}^{ax}}\sin \left( bx \right)}dx\]$

$I=\[\int{{{e}^{ax}}\sin \left( bx \right)}dx\]$  Eq1

Integrate using integration by parts

$\[I={{e}^{ax}}\int{\sin \left( bx\right)}dx-\int{\left( \frac{d\left( {{e}^{ax}} \right)}{dx}.\left( \int{\sin \left( bx \right)}dx \right) \right)}dx\]$

Use integration identity$\[\int{\sin \left( ax \right)dx=-\frac{\cos \left( ax \right)}{a}+c}\]$

$I={{e}^{ax}}\frac{-\cos \left( bx \right)}{b}-\int{\left( a{{e}^{ax}}.\left( \frac{-\cos \left( bx \right)}{b} \right) \right)}dx+c \\ & \Rightarrow I=-{{e}^{ax}}\frac{\cos \left( bx \right)}{b}+\frac{a}{b}\int{{{e}^{ax}}\cos \left( bx \right)}dx+c \\ \end{align}$

$\[\begin{align} & I={{e}^{ax}}\frac{-\cos \left( bx+c \right)}{b}-\int{\left( a{{e}^{ax}}.\left( \frac{-\cos \left( bx+c \right)}{b} \right) \right)}dx+c \\ & \Rightarrow I=-{{e}^{ax}}\frac{\cos \left( bx \right)}{b}+\frac{a}{b}\int{{{e}^{ax}}\cos \left( bx \right)}dx+c \\ \end{align}\]$

Integrate using integration by parts

$\[I=-{{e}^{ax}}\frac{\cos \left( bx \right)}{b}+\frac{a}{b}\left\{ {{e}^{ax}}\int{\cos \left( bx \right)}dx-\int{\left( \frac{d\left( {{e}^{ax}} \right)}{dx}.\left( \int{\cos \left( bx \right)}dx \right) \right)}dx \right\}+c\]$

Use integration identity$\[\int{\cos \left( ax \right)dx= \frac{\sin \left( ax \right)}{a}+c}\]$

$I=-{{e}^{ax}}\frac{\cos \left( bx \right)}{b}+\frac{a}{b}\left\{ {{e}^{ax}}\frac{\sin \left( bx \right)}{b}-\int{\left( a{{e}^{ax}}.\left( \frac{\sin \left( bx \right)}{b} \right) \right)}dx \right\}+C \\ & \Rightarrow I=-{{e}^{ax}}\frac{\cos \left( bx \right)}{b}+a{{e}^{ax}}\frac{\sin \left( bx \right)}{{{b}^{2}}}-\frac{{{a}^{2}}}{{{b}^{2}}}\int{{{e}^{ax}}\sin \left( bx \right)}dx+C \\ \end{align}$

Substitute from Eq1$I=\[\int{{{e}^{ax}}\sin \left( bx \right)}dx\]$

$\[I=-{{e}^{ax}}\frac{\cos \left( bx \right)}{b}+a{{e}^{ax}}\frac{\sin \left( bx \right)}{{{b}^{2}}}-\frac{{{a}^{2}}}{{{b}^{2}}}I+C\]$

Multiply both sides by b2 and take I on left side

${{b}^{2}}I=-{{e}^{ax}}b\cos \left( bx \right)+{{e}^{ax}}a\sin \left( bx \right)-{{a}^{2}}I+C \\ & \Rightarrow {{a}^{2}}I+{{b}^{2}}I=-{{e}^{ax}}b\cos \left( bx \right)+{{e}^{ax}}a\sin \left( bx \right)+C \\ & \Rightarrow \left( {{a}^{2}}+{{b}^{2}} \right)I={{e}^{ax}}\left( a\sin \left( bx \right)-b\cos \left( bx \right) \right) \\ \end{align}$

Divide both sides by  a² + b²

$\[I=\frac{{{e}^{ax}}}{{{a}^{2}}+{{b}^{2}}}\left( a\sin \left( bx \right)-b\cos \left( bx \right) \right)+C\]$

$\[\int{{{e}^{ax}}\sin \left( bx \right)}dx\]= \dfrac{{e}^{ax}}{{{a}^{2}}+{{b}^{2}}}\left( a\sin \left( bx \right)-b\cos \left( bx \right) \right)+C$

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Answer 2

Answer:

☬If a=b^x ,\ b=c^y and \ c=a^z , then the value of x y z is equal to <br> a. 0 <br> b. 1 <br> c. -1 <br> d. a b c.