How do I take this definite integral? FRom x^2 to sin(x), of t / t^4 + 1, dt.

**Gui** · June 14th 2012, 01:58 PM

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!

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

**a tutor** · June 14th 2012, 02:23 PM

The substitution u=t^2 works.

**Kmath** · June 14th 2012, 02:43 PM

Originally Posted by a tutor

The substitution u=t^2 works.

yes it works and you get $\arctan{t^2}|^{\sin x}_{x^2}$ . But, usually question like this ask to find the derivative of this term, do not it?

**Gui** · June 14th 2012, 02:48 PM

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.

**Kmath** · June 14th 2012, 02:56 PM

Originally Posted by Gui

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.

yes of course, but I neglected 1/2.

**Gui** · June 14th 2012, 02:58 PM

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!

How do I take this definite integral? FRom x^2 to sin(x), of t / t^4 + 1, dt.-image-19-.jpg

How do I take this definite integral? FRom x^2 to sin(x), of t / t^4 + 1, dt.-image-20-.jpg

**Kmath** · June 14th 2012, 03:05 PM

How did you deduce the last move of the first page?
Are you sure that the question need not the derivative?

**Gui** · June 14th 2012, 03:08 PM

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?

**Kmath** · June 14th 2012, 03:15 PM

[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.

**Kmath** · June 14th 2012, 03:21 PM

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 $(\frac{f}{g})^\prime=\frac{f^\prime g-fg^\prime}{g^2}$ .

**Gui** · June 14th 2012, 03:28 PM

[QUOTE=Kmath;723173]

Originally Posted by Gui

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.

Sorry, I didn't go over this in class. From reading the wiki page, I'm thinking it's the First Theorem I should be using? How would I apply it to the problem?

**Kmath** · June 14th 2012, 03:42 PM

[QUOTE=Gui;723177]

Originally Posted by Kmath

Sorry, I didn't go over this in class. From reading the wiki page, I'm thinking it's the First Theorem I should be using? How would I apply it to the problem?

Ok, I did not find what I meant in Wikipedia. Any way, here is roughly the Theorem

$\frac{d}{dx}\int_{f(x)}^{g(x)}h(t)dt=h(f(x))f^{\pr ime}(x)-h(g(x))g^{\prime}(x)$ .
So, in your case we have

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

You may find the theorem in any calculus book.

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