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Mathematics 15 Online
OpenStudy (anonymous):

integral of cos(lnx)dx

OpenStudy (anonymous):

\[\int\limits_{?}^{?}\cos(lnx)dx\]

OpenStudy (anonymous):

so u = lnx , du= 1/x dx

OpenStudy (anonymous):

You can put : cos(lnx) = u and x^0 = v

OpenStudy (anonymous):

\[\int\limits x^0 \cdot \cos(\ln(x)) \cdot dx\]

OpenStudy (anonymous):

Now you can use Integration by parts..

OpenStudy (anonymous):

what is du of cos(lnx)?

OpenStudy (anonymous):

Sorry, first function is : x^0

OpenStudy (anonymous):

so u is 1?

OpenStudy (anonymous):

yes..

OpenStudy (anonymous):

question why some time i have to do x^0 but some time i dont

OpenStudy (anonymous):

nvm to do integral by part i think

OpenStudy (anonymous):

Wait, I was right earlier..

OpenStudy (anonymous):

what is the anti derivation of cos(lnx)?

OpenStudy (anonymous):

Here, first function is : cos(lnx) And second function is : 1

OpenStudy (anonymous):

Try this now..

OpenStudy (anonymous):

so u is cos(lnx)?

OpenStudy (anonymous):

\[\frac{d}{dx} \cos(lnx) = -\frac{1}{x}\cdot \sin(lnx)\]

OpenStudy (anonymous):

Yes..

OpenStudy (anonymous):

\[\int\limits 1 \cdot dx = x\]

OpenStudy (anonymous):

ok i got it from now on thx you

OpenStudy (anonymous):

Note, you will have to use Parts twice here..

OpenStudy (anonymous):

k

OpenStudy (anonymous):

First by parts will give you: \[\int\limits \cos(lnx) \cdot dx = x \cdot \cos(lnx) + \int\limits \sin(lnx) \cdot dx\]

OpenStudy (anonymous):

Then again integral on right side, again first function = sin(lnx) and second function = 1. Then do the rest..

OpenStudy (anonymous):

i got it thx you

OpenStudy (anonymous):

You are welcome dear.. :)

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