A few integrals from my homework

[wetwilly92]

#1

Homework Statement

\int r^4 ( ln (r) ) dr

Homework Equations

Infinity algebra and Calc related formulae..

The Attempt at a Solution

Not sure even where to start here.. I'm thinking a u-substitution, letting u = r^3 so that I can deal with the two left over r's, but I don't think that it would be valid algebra.

EDIT: I Think I got this one using integration by parts, but I'd still like to make sure..

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#2

Homework Statement

\int 2t dt / (t^2 -6t + 9)

Homework Equations

Infinity algebra and Calc related formulae..

The Attempt at a Solution

I tried partial fraction decomposition here, but I kept getting nonsensical solutions. How do I set this one up?

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#3

Homework Statement

\int x sin^2 (x)

Homework Equations

Infinity algebra and Calc related formulae..

The Attempt at a Solution

Here I tried integration by parts, with

x(x-2sin(2x)) / 4 - 1/4 \int x - 2sin 2x dx + C which I also cannot integrate.

Thanks for any and all help!

 

[lurflurf]

first one integrate by parts

second one partial fraction decomposition works fine as does the substitution
u=t^2-6t+9 or u=t-3
for partial fraction decomposition chose A and B such that
2t=A+B(t-3)

 

[Dembadon]

#1

Homework Statement

\int r^4 ( ln (r) ) dr

Homework Equations

Infinity algebra and Calc related formulae..

The Attempt at a Solution

Not sure even where to start here.. I'm thinking a u-substitution, letting u = r^3 so that I can deal with the two left over r's, but I don't think that it would be valid algebra.

EDIT: I Think I got this one using integration by parts, but I'd still like to make sure..

[...]

I would let u = r⁴ for this problem.

 

[wetwilly92]

I would let u = r⁴ for this problem.

but don't we have to account for 3 more 'r's if we do this since du = r^3 dr?

 

[lineintegral1]

Okay, let's see if we can figure out what method to use with each. You are correct, you should use parts for the first integral. For the second integral, partial fractions will not work because the denominator is a square. Did you try a substitution? Perhaps try factoring then substituting.

I don't understand what the problem is with the last one. Your iteration of parts simply yields an elementary integral. Why can't you solve this?

 

[lurflurf]

^Partial fraction decomposition works fine, chose A and B such that
2t=A+B(t-3)

 

[Dembadon]

but don't we have to account for 3 more 'r's if we do this since du = r^3 dr?

Don't confuse substitution integration with parts integration.

When performing integration by parts, you want to choose u, such that u is easy to differentiate, and dv such that dv is easy to integrate. You're not looking to cancel-out any r in this problem.

 

[Char. Limit]

Don't confuse substitution integration with parts integration.

When performing integration by parts, you want to choose u, such that u is easy to differentiate, and dv such that dv is easy to integrate. You're not looking to cancel-out any r in this problem.

In that case, wouldn't we let u=ln(r), as ln(r) is easy to differentiate but hard to integrate?

I would have u=ln(r), not u=r^4.

 

[SammyS]

#1

\int r^4 (\ln (r) ) dr

EDIT: I Think I got this one using integration by parts, but I'd still like to make sure..
---
#2

\int 2t dt / (t^2 -6t + 9)

The Attempt at a Solution

I tried partial fraction decomposition here, but I kept getting nonsensical solutions. How do I set this one up?
---
#3

\int x sin^2 (x)

x(x-2sin(2x)) / 4 - 1/4 \int x - 2sin 2x dx + C which I also cannot integrate.

Thanks for any and all help!

For #1: Show your solution & answer so we can check it.

For #2. Yes, you can use partial fractions, but notice that the denominator is a perfect square.

(t^2 -6t + 9)=(x-3)^2

Use the substitution u = t-3. It works out very nicely.

For #3: Just finish it.

 

[Char. Limit]

For #2, add and subtract 6 from the numerator, like so:

\int \frac{2t-6+6}{t^2-6t+9} dt

Then split it into two parts:

\int \frac{2t-6}{t^2-6t+9} dt + \int \frac{6}{t^2-6t+9} dt

Both parts of that integral can be computed easily. I recommend u-substitution for the first one.

 

[Dembadon]

In that case, wouldn't we let u=ln(r), as ln(r) is easy to differentiate but hard to integrate?

I would have u=ln(r), not u=r^4.

You're absolutely right.

A good reason why I shouldn't be providing "help" while working! Sorry, wetwilly.