# Integration By Parts

∫cosx(lnsinx)dx

## 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?

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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 Helper

∫cosx(lnsinx)dx

## 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.

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