## anonymous 5 years ago integration

1. anonymous

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

2. anonymous

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

3. anonymous

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