Ryan Rigdon
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- Jun 10, 2010
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how does my work look?
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evaluating
- Thread starter Ryan Rigdon
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Check your answer by differentiating....
BigGlenntheHeavy
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\(\displaystyle Evaluate \ 7\int sin[ln|8x|]dx.\)
\(\displaystyle OK, \ let \ k \ = \ ln|8x|, \ dk \ = \ \frac{dx}{x}, \ and \ x \ = \ \frac{e^k}{8}.\)
\(\displaystyle Therefore, \ 7\int sin[ln|8x|]dx \ = \ \frac{7}{8}\int sin(k)e^kdk, \ remember \ this.\)
\(\displaystyle Now \ \frac{7}{8}\int sin(k)e^kdk, \ let \ u \ = \ sin(k), \ du \ = \ cos(k)dk\)
\(\displaystyle and \ dv \ = \ e^kdk, \ \implies \ v \ = \ e^k.\)
\(\displaystyle Hence, \ we \ have \ \frac{7}{8}\int sin(k)e^kdk \ = \ \frac{7}{8}e^ksin(k)-\frac{7}{8}\int e^kcos(k)dk.\)
\(\displaystyle Now, \ \int e^kcos(k)dk \ = \ e^kcos(k)+\int e^ksin(k)dk.\)
\(\displaystyle Ergo, \ we \ now \ have: \ \frac{7}{8}\int sin(k)e^kdk \ = \ \frac{7}{8}e^ksin(k)-\frac{7}{8}e^kcos(k)-\frac{7}{8}\int e^ksin(k)dk.\)
\(\displaystyle Hence, \ \frac{7}{4}\int e^ksin(k)dk \ = \ \frac{7}{8}e^ksin(k)-\frac{7}{8}e^kcos(k)+C \ or \ \int e^ksin(k)dk \ = \ \frac{e^ksin(k)}{2}-\frac{e^kcos(k)}{2}+C.\)
\(\displaystyle \int e^ksin(k)dk \ = \ 8\int sin[ln|8x|}dx \ = \ 4xsin[ln|8x|]-4xcos[ln|8x|] +C.\)
\(\displaystyle Therefore \ 7\int sin[ln|8x|]dx \ = \ \frac{7x}{2}[sin(ln|8x|)-cos(ln|8x|)]+C, \ QED.\)
\(\displaystyle Ryan, \ where \ do \ you \ get \ these \ "Icky" \ problems?\)
\(\displaystyle If \ these \ are \ the \ problems \ your \ Professor \ gives \ you, \ then \ he \ or \ she \ must \ indeed\)
\(\displaystyle be \ a \ stern \ taskmaster.\)
\(\displaystyle OK, \ let \ k \ = \ ln|8x|, \ dk \ = \ \frac{dx}{x}, \ and \ x \ = \ \frac{e^k}{8}.\)
\(\displaystyle Therefore, \ 7\int sin[ln|8x|]dx \ = \ \frac{7}{8}\int sin(k)e^kdk, \ remember \ this.\)
\(\displaystyle Now \ \frac{7}{8}\int sin(k)e^kdk, \ let \ u \ = \ sin(k), \ du \ = \ cos(k)dk\)
\(\displaystyle and \ dv \ = \ e^kdk, \ \implies \ v \ = \ e^k.\)
\(\displaystyle Hence, \ we \ have \ \frac{7}{8}\int sin(k)e^kdk \ = \ \frac{7}{8}e^ksin(k)-\frac{7}{8}\int e^kcos(k)dk.\)
\(\displaystyle Now, \ \int e^kcos(k)dk \ = \ e^kcos(k)+\int e^ksin(k)dk.\)
\(\displaystyle Ergo, \ we \ now \ have: \ \frac{7}{8}\int sin(k)e^kdk \ = \ \frac{7}{8}e^ksin(k)-\frac{7}{8}e^kcos(k)-\frac{7}{8}\int e^ksin(k)dk.\)
\(\displaystyle Hence, \ \frac{7}{4}\int e^ksin(k)dk \ = \ \frac{7}{8}e^ksin(k)-\frac{7}{8}e^kcos(k)+C \ or \ \int e^ksin(k)dk \ = \ \frac{e^ksin(k)}{2}-\frac{e^kcos(k)}{2}+C.\)
\(\displaystyle \int e^ksin(k)dk \ = \ 8\int sin[ln|8x|}dx \ = \ 4xsin[ln|8x|]-4xcos[ln|8x|] +C.\)
\(\displaystyle Therefore \ 7\int sin[ln|8x|]dx \ = \ \frac{7x}{2}[sin(ln|8x|)-cos(ln|8x|)]+C, \ QED.\)
\(\displaystyle Ryan, \ where \ do \ you \ get \ these \ "Icky" \ problems?\)
\(\displaystyle If \ these \ are \ the \ problems \ your \ Professor \ gives \ you, \ then \ he \ or \ she \ must \ indeed\)
\(\displaystyle be \ a \ stern \ taskmaster.\)