Struggling with Integration: A Call for Assistance

[dalterego]

Integration Issues...a bit urgent

Hey having problems with five of my assignment questions, any help would be appreciated.

Homework Statement

(problems attached)

Homework Equations

The Attempt at a Solution

For the first one, I tried to solve it by using the sin^-1 x as 'u' and integrating the u and getting u³ / 3 + c, but that doesn't seem to work. Then I thought I could use the Integration by parts formula, but since its (sin^-1x)², both u and v would be the same...

Second one: I thought of taking the y^0.5 as the substitution 'u', but it doesn't seem to make much difference.

Third one: I thought I could separate it into ln( (1+t)^0.5) + ln (t^0.5) and solve them individually...

Fourth one: The improper fractions technique, but then I can't seem to get legit values for A, B and C

Last one: Direct Comparison Test, comparing it to the fraction: x² / (x+3)³

 

[foxjwill]

Why would it matter if u and dv (not v) are the same? In fact, for 1, that would be probably the only way to do it.

 

[dynamicsolo]

I'll need to deal with this in parts. Here's what I suggest at the moment.

1) It may be six-of-one, half-a-dozen-of-the-other as to whether you should use u = (arcsin x) or u = (arcsin x)^2 .

If you do the former, you will have to integrate dv = (arcsin x) dx , which isn't too bad. It's tidier, though, to do the latter and have dv = dx.

It looks like either way you will have to integrate

[tex]\frac{x}{\sqrt{1 - x^2} } arcsin x dx [/tex]

which means a second integration by parts with u = arcsin x and dv for the rest, which is a substitution integral. I think it does settle down after that... (Nasty little problem.)

EDIT: Worked it out with u = (arcsin x)^2. The second integration-by-parts is easy.

2) Try v = sqrt(y). You'll then have an integration-by-parts to do; the v du integral looks like it will require a trig substitution, but then you're done.

4) Another multiple-technique problem. (I'm suspecting they all are: I think your instructor is out to get you...) Substitute u = e^x to create an integrand with a quadratic polynomial over the square root of a quadratic polynomial. Then try completing the square for the argument of the radical. It may be a trig-substitution integral from there...

5) I take it you're supposed to see if this improper integral diverges, or else evaluate it if it converges. You're correct in that you want to focus on the [tex]\frac{x^2}{(x+3)^3}[/tex] term. The (sin x)^2 factor is there, I suspect, to attempt to foil you in this, since it is bounded above by 1, so it is "usually" smaller than [tex]\frac{x^2}{(x+3)^3}[/tex]. You may have to make some sort of argument that, averaged over a period, sine-squared is non-zero, so this integral term acts like a constant times 1/x , and thus diverges. (One certainly has an intuition that this thing diverges...)

Still thinking about #3...

 

[foxjwill]

for 3, your best bet would probably be to use integration by parts with u=1 and dv=... . Doing so will "get rid" of the ln under the integral sign.

 

[gamesguru]

Ha, I remember doing the first four from my textbook.
1) Use parts, let u=arcsin[x]^2 dv=dx. Should be clear what to do next.
2) Use sub, let u=sqrt[x] so 2u du =dx, then use parts.
Never could solve the others, didn't really care either because they never come up in the real world.

 

[Defennder]

For 3, the solution I could come up with involves 2 substitutions and quite a number of integration by parts. As foxjwill said, do it first by integration by parts, then make a substitution for t^(1/2), you'll end up with something you should recognise easily as integrable by a trigo substitution, but it gets tedious from this point, though it's doable.

EDIT: The online integrator's answer involves a hyperbolic sine, which strongly implies that you can shorten the number of steps needed by familiarising yourself with integrals and differentiation of hyperbolic trigo functions.

 

[Gib Z]

That is correct, for the third one [itex] t = \sinh^2 x[/itex] is good to use, it comes out in a few lines if you are familiar with hyperbolic trig functions.