I try to find a reasonable solution for this equation but i couldent
I try to study lots of material but i couldent solve it. I am a high school student and try to learn. Integral cos(log x)dx
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See here. You and the other user may be in the same contest ... – user138999 Jul 12 '14 at 02:42
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4@J.Finnegan It's the same user who asked both questions. – Jul 12 '14 at 02:43
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Its not duplicate its log x my previous question is ln x – Nicklaus Josef Jul 12 '14 at 02:44
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$\log(x)$ is often assumed to have base $e$, what base are you using? – user138999 Jul 12 '14 at 02:45
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1Considering that $\log_{10} x = \ln x / \ln 10$, surely you can use the answer to the previous question, after a trivial change of variables. – Jul 12 '14 at 02:45
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Edit It looks from later comments that you may be interested in log to the base $10$. Whatever base $b$ you are interested in, there is an easily computed constant $a$ such that $\log_b(x)=a\ln x$. So we integrate $\cos(a\ln x)$.
We try integration by parts, $u=\cos(a\ln x)$ and $dv=dx$. Then $du=-a\frac{1}{x}\sin(a\ln x)$ and we can take $v=x$. Thus our integral is $$x\cos(a\ln x)+a\int \sin(a\ln x)\,dx.$$ Now attack the second integral. The same basic strategy shows that $$a\int \sin(a\ln x)\,dx=ax\sin(a\ln x) -a^2\int \cos(a\ln x)\,dx.$$
It looks as if we are going in circles. And usually when it looks as if we are going in circles, we are going in circles. But not this time.
Let $I$ be our original integral. Then $$I=x\cos(a\ln x)+ ax\sin(a\ln x) -a^2I.$$ Solve for $I$, and don't forget the constant of integration.
André Nicolas
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Duplicate of solution http://math.stackexchange.com/a/864776/123905 at duplicate of this question, asked by the same user. – Eric Towers Jul 12 '14 at 02:54
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@EricTowers This is not a "duplicate answer." Accusing someone of writing a "duplicate question" is minor--that just means the person may not have searched long enough. Accusing someone of writing a "duplicate answer" is essentially accusing them of plagiarism. In order to do so, you better have some pretty strong proof. We don't consider something to be a duplicate answer unless it is a character-for-character copy (which this is not). – apnorton Jul 12 '14 at 02:56
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1@anorton: Question re-posed by same questioner, commented duplicate 7-8 minutes prior to this answer being posted. There is nothing here that is not in the prior answer. Value to the site of duplication of answers: zero. – Eric Towers Jul 12 '14 at 03:03
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1@EricTowers Do you mean "asker"? (I was specifically looking to see if Andre had posted in the other thread, and I didn't see him.) I grant that the value to the site of duplicate answers is zero, but the designed mechanism to handle this is closure of the question (not punishing the answerer). If it were the responsibility of an answerer to check for a duplicate answer, there would be many fewer questions and answers here. – apnorton Jul 12 '14 at 03:07
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1@anorton: I observe that questioner's third integral in an hour ( http://math.stackexchange.com/questions/864794/sin-x-integral-qestions ) has a reference to one of the times that André Nicolas has previously addressed it. I don't plan any accusations based on this. :-) – Eric Towers Jul 12 '14 at 03:17