Integration Problem
- Thread starter sunrise
- Start date
Hello, sunrise!
This requires Integration By Parts . . . twice.
By parts: .\(\displaystyle \begin{Bmatrix}u &=& \sin(bx+c) && dv &=& e^{ax}\,dx \\ du &=& b\cos(bx+c)\,dx && v &=& \frac{1}{a}e^{ax} \end{Bmatrix}\)
Then: .\(\displaystyle \displaystyle I \;=\;\tfrac{1}{a}e^{ax}\sin(bx+c) - \tfrac{b}{a}\int e^{ax}\cos(bx+c)\,dx \)
By parts: .\(\displaystyle \begin{Bmatrix}u &=& \cos(bx+c) && dv &=& e^{ax}x \\ du &=& \text{-}b\sin(bx+c) && v &=& \frac{1}{a}e^{ax} \end{Bmatrix}\)
Then: .\(\displaystyle \displaystyle I \;=\;\tfrac{1}{a}e^{ax}\sin(bx+c) - \tfrac{b}{a}\left[\tfrac{1}{a}\cos(bx+c) + \tfrac{b}{a}\int e^{ax}\sin(bx+c)\,dx\right] \)
. . . . . \(\displaystyle \displaystyle I \;=\;\tfrac{1}{a}e^{ax}\sin(bx+c) - \tfrac{b}{a^2}e^{ax}\cos(bx+c) - \tfrac{b^2}{a^2}\underbrace{\int e^{ax}\sin(bx+c)\,dx}_{\text{This is }I} \)
. . . . . \(\displaystyle I \;=\;\frac{1}{a}e^{ax}\sin(bx+c) - \frac{b}{a^2}e^{ax}\cos(bx+c) - \frac{b^2}{a^2}I + C\)
Multiply by \(\displaystyle a^2\!:\)
. . . . . . . \(\displaystyle a^2I \;=\;ae^{ax}\sin(bx+c) - be^{ax}\cos(bx+c) - b^2I +C\)
. . .\(\displaystyle a^2I + b^2I \;=\;ae^{ax}\sin(bx+c) - be^{ax}\cos(bx+c) + C \)
. . \(\displaystyle (a^2+b^2)I \;=\;e^{ax}\big[a\sin(bx+c) - b\cos(bx+c)\big] + C\)
. . . . . . . . .\(\displaystyle I \;=\;\dfrac{e^{ax}}{a^2+b^2}\big[a\sin(bx+c) - b\cos(bx+c)\big] + C\)
This requires Integration By Parts . . . twice.
By parts: .\(\displaystyle \begin{Bmatrix}u &=& \sin(bx+c) && dv &=& e^{ax}\,dx \\ du &=& b\cos(bx+c)\,dx && v &=& \frac{1}{a}e^{ax} \end{Bmatrix}\)
Then: .\(\displaystyle \displaystyle I \;=\;\tfrac{1}{a}e^{ax}\sin(bx+c) - \tfrac{b}{a}\int e^{ax}\cos(bx+c)\,dx \)
By parts: .\(\displaystyle \begin{Bmatrix}u &=& \cos(bx+c) && dv &=& e^{ax}x \\ du &=& \text{-}b\sin(bx+c) && v &=& \frac{1}{a}e^{ax} \end{Bmatrix}\)
Then: .\(\displaystyle \displaystyle I \;=\;\tfrac{1}{a}e^{ax}\sin(bx+c) - \tfrac{b}{a}\left[\tfrac{1}{a}\cos(bx+c) + \tfrac{b}{a}\int e^{ax}\sin(bx+c)\,dx\right] \)
. . . . . \(\displaystyle \displaystyle I \;=\;\tfrac{1}{a}e^{ax}\sin(bx+c) - \tfrac{b}{a^2}e^{ax}\cos(bx+c) - \tfrac{b^2}{a^2}\underbrace{\int e^{ax}\sin(bx+c)\,dx}_{\text{This is }I} \)
. . . . . \(\displaystyle I \;=\;\frac{1}{a}e^{ax}\sin(bx+c) - \frac{b}{a^2}e^{ax}\cos(bx+c) - \frac{b^2}{a^2}I + C\)
Multiply by \(\displaystyle a^2\!:\)
. . . . . . . \(\displaystyle a^2I \;=\;ae^{ax}\sin(bx+c) - be^{ax}\cos(bx+c) - b^2I +C\)
. . .\(\displaystyle a^2I + b^2I \;=\;ae^{ax}\sin(bx+c) - be^{ax}\cos(bx+c) + C \)
. . \(\displaystyle (a^2+b^2)I \;=\;e^{ax}\big[a\sin(bx+c) - b\cos(bx+c)\big] + C\)
. . . . . . . . .\(\displaystyle I \;=\;\dfrac{e^{ax}}{a^2+b^2}\big[a\sin(bx+c) - b\cos(bx+c)\big] + C\)
D
Deleted member 4993
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In every text-book that I know of solves a very similar problem in their example-set:
[FONT=MathJax_Math]I [/FONT][FONT=MathJax_Main]= [/FONT][FONT=MathJax_Size2]∫[/FONT][FONT=MathJax_Math]e[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Main](x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]x[/FONT]
It should be few simple steps from there to the actual problem.
