Mathematics
15 Online
OpenStudy (anonymous):
integral of cos(lnx)dx
Join the QuestionCove community and study together with friends!
Sign Up
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..
Join the QuestionCove community and study together with friends!
Sign Up
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
Join the QuestionCove community and study together with friends!
Sign Up
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..
Join the QuestionCove community and study together with friends!
Sign Up
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
Join the QuestionCove community and study together with friends!
Sign Up
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
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
You are welcome dear.. :)