Question

integration

Answer 1

\[\int\limits_{}^{}\cos (b \log_{e}(x/a) )dx\]

Answer 2

\[\int\limits \cos[bln(\frac{x}{a})]dx \rightarrow u=\ln(\frac{x}{a}), xdu=dx \rightarrow \int\limits xcos[bln(u)]du\]
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)...
\[\int\limits \cos[bln(\frac{x}{a})]dx= \frac{xcos[bln(\frac{x}{a})]+bxsin[bln(\frac{x}{a})]}{1+b^2}+C\]

Answer 3

Sorry made a typo in the u sub, it should be this...
\[∫\cos[bln(\frac{x}{a})]dx→u=\ln(\frac{x}{a}),xdu=dx→∫xcos[bu]du\]