# Evaluate the definite integral!

1. Mar 20, 2005

### EasyStyle4747

There are several situations I don't know how to solve:

integral of (lnx)/(x^2) dx

integral of e^x(cos2x)dx

integral of ln(1+x^2)dx

I have a quiz tommorow, and after looking through some notes, this is what I couldn't understand.

Please help! thanks.

2. Mar 20, 2005

### hypermorphism

These can all be done by the appealing to the product rule (integration by parts). Show your work so that we can help with where you're stuck.

3. Mar 20, 2005

### Jameson

$$\int udv = uv - \int{vdu}$$

4. Mar 20, 2005

### EasyStyle4747

Yes, I realize its integration by parts. I've been doing other problems with the product rule. Its just these that I'm stumped on. I basically don't know where to begin with some of these.

for this one: integral of (lnx)/(x^2) dx , are you supposed to change it to (lnx)(x^-2) first and then use parts?

for this one: ntegral of e^x(cos2x)dx, o jeez, i forgot, how do you take the anti-derivative of cos2x :surprised

for this one: integral of ln(1+x^2)dx , how do you split it into 2 parts? What do you set u equal to? What do you set dv equal to?

Some general questions:
-Does it matter which part you set dv or u equal to? Or do you HAVE to go in order?

thnx for being helpful people!

5. Mar 21, 2005

### t!m

Order does in fact matter, and generally LIATE (or ILATE) will help you choose what to make 'u' : Logarithmic, Inverse, Algebraic, Trig, Exponential. For example, your first problem:

$$\int \frac{lnx}{x^2}dx$$

let $$u=lnx; dv=\frac{1}{x^2}dx$$
And just derive/integrate to find du and v and plug in to the aforementioned expression.

6. Mar 21, 2005

### dextercioby

Those integrals:

a) are not definite.Are indefinite,equivalently,are antiderivarives of certain functions.
b)can be computed via part integration.For the second,two times partial integration is necessary.

7. Mar 21, 2005

### Nylex

The integral of cos 2x = (1/2)sin 2x.

You can write that as 1.ln(1 + x^2) dx, but there's probably a better way of doing that integral.

Yeah, it does because one way round you probably won't be able to do the integral. Usually, you take u to be the part that is "easier" to deal with (eg. polynomials, like x^2) and the dv shouldn't get "harder" to deal with when you integrate (eg. exponentials, trig functions). Say you had the integral of x^2.e^x, you'd take u = x^2 and dv/dx = e^x (and you'd have to do integration by parts twice).

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