# Problem with integration

• Mar 29th 2009, 07:53 AM
collegestudent321
Problem with integration
Hello, I am currently working on a problem that requires the integration of an expression that I can't seem to figure out. The problem goes:

Integral from 0 to 2 of: (((t^2)+2t+2)/((1+t)^4))^1/2

So far, I know that I have to use u-substitution and I set u = 1+t, so du = dt. Now, i have

Integral of: ((u^2 + 1)^1/2)/(u^2)

• Mar 29th 2009, 08:00 AM
Jester
Quote:

Originally Posted by collegestudent321
Hello, I am currently working on a problem that requires the integration of an expression that I can't seem to figure out. The problem goes:

Integral from 0 to 2 of: (((t^2)+2t+2)/((1+t)^4))^1/2

So far, I know that I have to use u-substitution and I set u = 1+t, so du = dt. Now, i have

Integral of: ((u^2 + 1)^1/2)/(u^2)

Try $\displaystyle u = \tan \theta$.
• Mar 29th 2009, 08:01 AM
collegestudent321
so use u = tan (t) instead of (1+t)??
• Mar 29th 2009, 08:06 AM
Jester
Quote:

Originally Posted by collegestudent321
so use u = tan (t) instead of (1+t)??

No No. Your first substitution is

$\displaystyle t + 1 = u$, your second is $\displaystyle u = \tan \theta$.
• Mar 29th 2009, 08:12 AM
collegestudent321
ok, so now i have that u = tan(theta)
and now it is:
integral of : ((tan(theta)^2)+1)^1/2) / (tan(theta)^2) which equals:

integral of: sec(theta) / ((sec(theta)^2) - 1)

is that right so far? and if it is, where do i go now?
• Mar 29th 2009, 08:32 AM
Jester
Quote:

Originally Posted by collegestudent321
ok, so now i have that u = tan(theta)
and now it is:
integral of : ((tan(theta)^2)+1)^1/2) / (tan(theta)^2) which equals:

integral of: sec(theta) / ((sec(theta)^2) - 1)

is that right so far? and if it is, where do i go now?

I think you're missing a $\displaystyle du$ however with the tan subs., it's going to be a long process. Let's try another substitution.

$\displaystyle u = \sinh \theta$
• Mar 29th 2009, 08:47 AM
collegestudent321
ok, now i'm completely lost. I tried the same process as before and now i have:

integral of: cosh(t) / sinh(t)^2 dt

where do i go from here?
• Mar 29th 2009, 08:52 AM
collegestudent321
wait, can this be equal to :

integral of : coth(t)*csch(t) ???

and then use integration by parts?
• Mar 29th 2009, 09:02 AM
Jester
OK. Let me put in some steps to help clear this up.If $\displaystyle \int \frac{\sqrt{u^2+1}}{u^2}\, du$ if $\displaystyle u = \sinh \theta$ then $\displaystyle \int \frac{\sqrt{\sinh^2 \theta+1}}{\sinh^2 \theta}\, \cosh \theta d \theta = \int \frac{\cosh^2 \theta}{\sinh^2 \theta} d \theta$

so
$\displaystyle \int \frac{\sinh^2 \theta +1}{\sinh^2 \theta} d \theta = \int ( 1 + \text{csch}^2 \theta ) d \theta = \theta - \coth \theta + c$. Now backsubstitute.
• Mar 29th 2009, 09:15 AM
collegestudent321
OHHHHHHH. wow i got it. the answer is:

ln (u + ((u^2)+1)) - ((1-(u^2))^1/2)/(u)

(and then substitute u with (t+1)

Thank you for your help. I REALLY appreciate it.
• Mar 29th 2009, 09:27 AM
collegestudent321
calculus with parametric curves
actually, if you don't mind, there is one other problem that i've been having. I'm sorry to bug you so much.

I have to prove that if x = asin(theta) and y = bcos(theta), show that the total length of the ellipse is:

L = 4a * integral (from 0 to pi/2) of: 1 - ((e^2)sin(theta)^2))^1/2 (d(theta))

I am almost there, but i came across this part:

L = integral of: (a^2) - (c^2)(sin(theta)^2) and i'm not sure how to make the connection to find e^2. Any ideas????
• Mar 29th 2009, 09:49 AM
Jester
Quote:

Originally Posted by collegestudent321
actually, if you don't mind, there is one other problem that i've been having. I'm sorry to bug you so much.

I have to prove that if x = asin(theta) and y = bcos(theta), show that the total length of the ellipse is:

L = 4a * integral (from 0 to pi/2) of: 1 - ((e^2)sin(theta)^2))^1/2 (d(theta))

I am almost there, but i came across this part:

L = integral of: (a^2) - (c^2)(sin(theta)^2) and i'm not sure how to make the connection to find e^2. Any ideas????

$\displaystyle L = 4 \int_0^{\pi/2} \sqrt{x'^2+y'^2} d \theta = 4\int_0^{\pi/2} \sqrt{ a^2 \cos^2 \theta + b^2 \sin^2 \theta} d \theta =$

$\displaystyle 4a\int_0^{\pi/2} \sqrt{\cos^2 \theta + \frac{b^2}{a^2} \sin^2 \theta} d \theta = 4a\int_0^{\pi/2} \sqrt{1 - \left(1 - \frac{b^2}{a^2}\right) \sin^2 \theta} d \theta$

Now $\displaystyle b^2 = a^2 - c^2$ and $\displaystyle e = \frac{c}{a}$ so using these we get

$\displaystyle 4a\int_0^{\pi/2} \sqrt{1 - \left(1 - \frac{b^2}{a^2}\right) \sin^2 \theta} d \theta = 4a\int_0^{\pi/2} \sqrt{1 - e^2\sin^2 \theta} d \theta$
• Mar 29th 2009, 10:07 AM
collegestudent321
oh, so was i supposed to know that e = c/a or is that derived from the problem???
• Mar 29th 2009, 11:24 AM
Jester
Quote:

Originally Posted by collegestudent321
oh, so was i supposed to know that e = c/a or is that derived from the problem???

The eccentricity e is usually defined that way - have you not seen it?
• Mar 29th 2009, 12:16 PM
collegestudent321
ohhh, ok. haha wow, i was thinking the other e. ok, that makes sense, yea i have seen that before, i just hadn't applied it. thank you!!!