Hey, I'd appreciate any help taking the integral below. The two methods I know are by substitution and by parts, can I use either of those? Thanks in advance!

$\displaystyle \int_{x^2}^{sin(x)} \frac{t}{t^4 + 1}dt$

- Jun 14th 2012, 01:58 PMGuiHow do I take this definite integral? From x^2 to sin(x), of t / t^4 + 1, dt.
Hey, I'd appreciate any help taking the integral below. The two methods I know are by substitution and by parts, can I use either of those? Thanks in advance!

$\displaystyle \int_{x^2}^{sin(x)} \frac{t}{t^4 + 1}dt$ - Jun 14th 2012, 02:23 PMa tutorRe: How do I take this definite integral? FRom x^2 to sin(x), of t / t^4 + 1, dt.
The substitution u=t^2 works.

- Jun 14th 2012, 02:43 PMKmathRe: How do I take this definite integral? FRom x^2 to sin(x), of t / t^4 + 1, dt.
- Jun 14th 2012, 02:48 PMGuiRe: How do I take this definite integral? FRom x^2 to sin(x), of t / t^4 + 1, dt.
a tutor: I'm trying, but stuck. I'll try and get a scan of my results up.

Kmath: Isn't it arctan(t^2)/2? I put it into my ti-89 and it gave me that; wolframalpha times out. - Jun 14th 2012, 02:56 PMKmathRe: How do I take this definite integral? FRom x^2 to sin(x), of t / t^4 + 1, dt.
- Jun 14th 2012, 02:58 PMGuiRe: How do I take this definite integral? FRom x^2 to sin(x), of t / t^4 + 1, dt.
Ah ok :) Here's my work so far, but I'm pretty much stuck here... I don't know what to do with the logs. Thanks in advance for all your help!

Attachment 24081

Attachment 24082 - Jun 14th 2012, 03:05 PMKmathRe: How do I take this definite integral? FRom x^2 to sin(x), of t / t^4 + 1, dt.
How did you deduce the last move of the first page?

Are you sure that the question need not the derivative? - Jun 14th 2012, 03:08 PMGuiRe: How do I take this definite integral? FRom x^2 to sin(x), of t / t^4 + 1, dt.
Kmath: Yeah, the question does ask for the derivative. Shouldn't I actually find f(x) though, before finding f´(x)?

The last move of the first page.... well, the derivative of ln(x) is 1/x, so the antiderivative of 1/(u^2 + 1) is ln(u^2 + 1)/derivative of u^2 +1, which is ln(u^2 + 1)/2u. Is that not correct? - Jun 14th 2012, 03:15 PMKmathRe: How do I take this definite integral? FRom x^2 to sin(x), of t / t^4 + 1, dt.
[QUOTE=Gui;723170]Kmath: Yeah, the question does ask for the derivative. Shouldn't I actually find f(x) though, before finding f´(x)?

So you can find the derivative with out finding f(x). In this case you can use the fundamental theorem of calculus. - Jun 14th 2012, 03:21 PMKmathRe: How do I take this definite integral? FRom x^2 to sin(x), of t / t^4 + 1, dt.
The last move of the first page.... well, the derivative of ln(x) is 1/x, so the antiderivative of 1/(u^2 + 1) is ln(u^2 + 1)/derivative of u^2 +1, which is ln(u^2 + 1)/2u. Is that not correct?[/QUOTE]

May be you forgot to use the "Quotient rule" that is $\displaystyle (\frac{f}{g})^\prime=\frac{f^\prime g-fg^\prime}{g^2}$. - Jun 14th 2012, 03:28 PMGuiRe: How do I take this definite integral? FRom x^2 to sin(x), of t / t^4 + 1, dt.
- Jun 14th 2012, 03:42 PMKmathRe: How do I take this definite integral? FRom x^2 to sin(x), of t / t^4 + 1, dt.
[QUOTE=Gui;723177]Ok, I did not find what I meant in Wikipedia. Any way, here is roughly the Theorem

$\displaystyle \frac{d}{dx}\int_{f(x)}^{g(x)}h(t)dt=h(f(x))f^{\pr ime}(x)-h(g(x))g^{\prime}(x)$.

So, in your case we have

$\displaystyle \frac{\sin{x}}{1+{\sin^4 x}}\cos x-\frac{x^2}{1+x^8}(2x)$.

You may find the theorem in any calculus book.