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anonymous
 5 years ago
integration
anonymous
 5 years ago
integration

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{}\cos (b \log_{e}(x/a) )dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits \cos[bln(\frac{x}{a})]dx \rightarrow u=\ln(\frac{x}{a}), xdu=dx \rightarrow \int\limits xcos[bln(u)]du\] since... \[u=\ln(\frac{x}{a}), e^u=e^{\ln(\frac{x}{a})}, e^u=\frac{x}{a}, ae^u=x \rightarrow \int\limits ae^ucos(bu)du\] Now use integration by parts..... \[\int\limits ae^ucos(bu)du=...........\] \[f=ae^u, F=ae^u, g=\cos(bu), g'=bsin(bu)\] \[\rightarrow ae^ucos(bu)+ \int\limits abe^usin(bu)du\] integrate by parts once again.... \[f=abe^u, F=abe^u, g=\sin(bu), g'=bcos(bu)\] \[ae^ucos(bu)+abe^usin(bu) \int\limits ab^2e^ucos(bu)du\] In other words.... \[\int\limits ae^ucos(bu)du= ae^ucos(bu)+abe^usin(bu) \int\limits ab^2e^ucos(bu)du\] \[\int\limits ae^ucos(bu)du+ \int\limits ab^2e^ucos(bu)du= ae^ucos(bu)+abe^usin(bu)\] combine like terms and factor out a 1+b^2 \[(1+b^2) \int\limits ae^ucos(bu)du= ae^ucos(bu)+abe^usin(bu)\] Thus.... \[\int\limits ae^ucos(bu)du= \frac{ ae^ucos(bu)+abe^usin(bu)}{1+b^2} +C\] now substitue u back in u=ln(x/a)... \[\int\limits \cos[bln(\frac{x}{a})]dx= \frac{xcos[bln(\frac{x}{a})]+bxsin[bln(\frac{x}{a})]}{1+b^2}+C\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry made a typo in the u sub, it should be this... \[∫\cos[bln(\frac{x}{a})]dx→u=\ln(\frac{x}{a}),xdu=dx→∫xcos[bu]du\]
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