Question

integrate it:
cos(ln x)dx

Answer 1

\[u=\ln x\]\[du=\frac1xdx\implies dx=e^udu\]

Answer 2

We have
\[\int \cos (\ln x) dx\]
let
\[x= e^t\]
\[dx= e^t dt\]
so we get
\[\int \cos ( \ln(e^t) e^t dt\]
we get now
\[ \int \cos t \times e^t dt\]
Can you solve it now?

Answer 3

yes yes thanks

Answer 4

Great:D

Answer 5

By applying by parts
\[\cos(lnx)*x-\int\limits(x)(-\sin(lnx)*1/x)dx\]

Answer 6

considering cos(lnx) as 1st func
and dx as 2nd func

Answer 7

As I=∫Cos(lnx)dx
considering cos(lnx) as 1st func
and dx as 2nd func
By applying by parts

I=cos(lnx)∗x−∫(x)(−sin(lnx)∗1/x)dx
I=cos(lnx)∗x+∫(sin(lnx))dx

now again applying by parts on ∫(sin(lnx))dx
considering sin(lnx) as 1st func
and dx as 2nd func
I=cos(lnx)∗x+{sin(lnx)*x-∫x*cos(lnx)*1/xdx}
I=cos(lnx)∗x+{sin(lnx)*x-∫cos(lnx)dx}
I=cos(lnx)∗x+sin(lnx)*x-I
2I=cos(lnx)∗x+sin(lnx)*x+c
I={cos(lnx)∗x}/2+{sin(lnx)*x}/2+c/2
I={cos(lnx)∗x}/2+{sin(lnx)*x}/2+c'

Answer 8

So simple isn't it?..
:)

Answer 9

thanks

Answer 10

My pleasure