Integration by Parts: Solving ∫cosx(lnsinx)dx

[jdawg]

Homework Statement

∫cosx(lnsinx)dx

Homework Equations

The Attempt at a Solution

u=lnsinx dv=cosxdx
du=cosx/sinx dx v=sinx

=(lnsinx)(sinx)-∫(sinx)(cosx/sinx)dx
=(lnsinx)(sinx)-(sinx)+C

I thought that I did this correctly, but my teacher said that u should equal sinx. Why would u not equal lnsinx?

 

[jackarms]

If you differentiate, you get the original function, so you did it correctly. I think your teacher was saying u should equal sinx if you're doing a simple u-sub to integrate, since that method works as well. You couldn't have u be sinx for an integration by parts, since it isn't a complete term in the integrand -- it's only part of the natural log.

 

[Dick]

Homework Statement

∫cosx(lnsinx)dx

Homework Equations

The Attempt at a Solution

u=lnsinx dv=cosxdx
du=cosx/sinx dx v=sinx

=(lnsinx)(sinx)-∫(sinx)(cosx/sinx)dx
=(lnsinx)(sinx)-(sinx)+C

I thought that I did this correctly, but my teacher said that u should equal sinx. Why would u not equal lnsinx?

You did do it correctly. I think your teacher might be suggesting you do a u-substitution first and then integrate log(u) by parts. I think that's actually a little more complicated, not easier.

 

[jdawg]

If you differentiate, you get the original function, so you did it correctly. I think your teacher was saying u should equal sinx if you're doing a simple u-sub to integrate, since that method works as well. You couldn't have u be sinx for an integration by parts, since it isn't a complete term in the integrand -- it's only part of the natural log.

Ohh ok! I didn't know you could use u substitution on that one. Thanks for clearing that up :)

 

[Dick]

Ohh ok! I didn't know you could use u substitution on that one. Thanks for clearing that up :)

You can do a u substitution, but then you are left with log(u), which you then need to integrate by parts. Unless you've memorized the integral of log(u). I think your way is better.