Integrals: Solving with Substitutions

[Jacobpm64]

Find the integral.

http://img83.imageshack.us/img83/1228/int203bd.gif

This one is just so confusing. *sigh*

Find the antiderivative.

http://img83.imageshack.us/img83/7179/int264ol.gif

I don't know how to approach this one.. I'm guessing making some substitutions.. but i don't know how you actually work it when you make subtitutions.. just like in my other post.. I can make substitutions.. but i don't know what to do after that. I need one with substitutions worked for me if they're all similar.

 

[dav2008]

Would it help you to see how a substitution works if I worked out a simple problem?

\int\frac{2x}{\sqrt{x^2+1}} \ dx

u=x^2+1
du=2x\dx

\int\frac{1}{\sqrt{u}} \ du <-Substitute values for u and du as appropriate.

\int u^{-\frac{1}{2}} du<-Just rewriting the square root sign as a power of -1/2 to make it easier to see the integration.

2u^{\frac{1}{2}}

2(x^2+1)^{\frac{1}{2}}<--substitute back for u=x²

2\sqrt{x^2+1}

Take a minute to understand why the substitution worked. You want to put everything in terms of one variable. By choosing u to be the value in the square root you obtain a value of du that matches the other x and dx values.

 

[Jacobpm64]

How's this for the first one?

http://img97.imageshack.us/img97/7520/inttry206vz.gif

And this for the second?

http://img186.imageshack.us/img186/6919/noworkint266iv.gif

Another one came out with an undefined answer, but the answer in the back of the book turned it into natural logs.. hmm.. i don't know how that works.. But following the same pattern.. i'd get..

http://img88.imageshack.us/img88/1598/workint262dh.gif

hmm?

 

[dav2008]

Well see du is the differential of u...

If u is x^2+2x+2 then du would be (2x+2)dx

 

[Jacobpm64]

lol forget i said that..

ok, i understand how to get what du is equal to.. what i don't get now is.. in your example problem, you never substituted back for the value of du.. so would it even change my answer?

 

[pocoman]

In the second integral: u=1-4x so du=-4dx NOT 2dx.
and \int u^{-1} du = lnu NOT u^{-1}/0

 

[Nimz]

Look a bit more carefully at dav2008's example. Do you see du anywhere in the expression: 2u^1/2? That's why he never substituted the value of du back in.