- #1

- 98

- 0

## Homework Statement

let u=lnx

du=1/x*dx

dv=cosdx

v=-sin

## Homework Equations

Now im confused as im getting nowhere with this substiution, i learned the LIPTE rule but its quite confusing, i have a function within a function

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Cudi1
- Start date

- #1

- 98

- 0

let u=lnx

du=1/x*dx

dv=cosdx

v=-sin

Now im confused as im getting nowhere with this substiution, i learned the LIPTE rule but its quite confusing, i have a function within a function

- #2

- 98

- 0

- #3

- 2,967

- 5

[tex]

\int{\cos{(\ln{(x)})} \, dx}

[/tex]

If it is, make the substitution:

[tex]

t = \ln{(x)} \Rightarrow x = e^{t}

[/tex]

and substitute everywhere. The integral that you will get can be integrated by using integration by parts twice, or, if you know complex numbers, by representing the trigonometric function through the complex exponential.

- #4

- 98

- 0

it is just cos(lnx)dx

- #5

- 2,967

- 5

then proceed as I told you.

- #6

- 98

- 0

ok im getting an integral of the form : coste^tdt. is this correct?

- #7

- 2,967

- 5

yes. proceed by integration by parts or using complex exponentials.

- #8

- 98

- 0

ok thank you for the help, quick question why do we have to let t=lnx?

- #9

- 2,967

- 5

- #10

- 98

- 0

which leaves me with : 1/2x(cos(lnx)-sin(lnx)+c?

- #11

- 2,967

- 5

[tex]

\frac{1}{2} x \left[\cos{\left(\ln{(x)}\right)} + \sin{\left(\ln{(x)}\right)}\right]+ C

[/tex]

- #12

- 98

- 0

- #13

- 2,967

- 5

yes, the x's will cancel.

Share:

- Replies
- 3

- Views
- 2K