please help... Integrate cos(lnx)dx

You have to know one of the "special" integral rules. http://www.wolframalpha.com/input/?i=%20Integrate%20cos(lnx)dx&t=crmtb01

do you know integration by parts?

hello?

this is a doable integral...

you can use integration by parts to solve this, recall\[\int\limits u dv = uv - \int\limits v du\]

dpalnc yeah I do and I have done it by parts but I thought maybe there is another way u knw

\[\int\limits \cos(lnx) = \sin(\ln x) x \ln x- x\]

Let I = ∫ cos ( lnx) dx u = cos ( lnx) --> du = - sin ( lnx)/x ... I = xcos (lnx) + xsin (lnx) - I I = (1/2) x cos ( lnx ) + x sin ( lnx) + C

I tried that Chlorophyl but ddnt work because I cant go back if I derive now

Try this: http://www.intmath.com/blog/tanzalin-method-for-easier-integration-by-parts/4339 It is a different easier integration by parts if you are having trouble with that.

Apply tabular method, it'll be simple!

u=cos(ln x), du= -sin(ln x), dv=dx, v=x \[\int\limits \cos (\ln x) dx=x \cos(\ln x)-\int\limits x(-\sin(\ln x)\]\[=xcos(\ln x)+x \sin(\ln x)-\int\limits \cos(\ln x) dx \] w= sin (ln x), dw= cos (ln x), dz=dx, z=x \[\int\limits \cos(\ln x) dx= x \cos(\ln x) + wz-\int\limits z dw\]\[=x \cos(\ln x) + x\sin(\ln x)- \int\limits \cos(\ln x) dx\] therefore:\[\int\limits \cos(\ln x) dx= 1/2 (x \cos(\ln x)+ x \sin(\ln x))\]

The first time I tackled this integral I tried the substitution: u = ln(x) and got lost; I didnt see that I needed \[x=e^{u}\] however with the help of the internet I finally had the insight! u=ln(x),du=1/x dx\[x=e^{u}, e^{u}du=dx \]\[\int\limits e^{u} \cos(u) du = e^{u} \cos(u)+\int\limits e^{u}\sin(u) du\]\[= e^{u}\cos(u)+e^{u}\sin(u)-\int\limits e^{u}\cos(u) du\]\[2int e^{u}\cos(u) du= e^{u}\cos(u)+e^{u}\sin(u)\]substituting back we get \[\int\limits \cos(\ln x) dx= 1/2(x \cos(\ln x)+ x \sin(\ln x))\]

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