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!
$\displaystyle \int_{x^2}^{sin(x)} \frac{t}{t^4 + 1}dt$
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!
$\displaystyle \int_{x^2}^{sin(x)} \frac{t}{t^4 + 1}dt$
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: 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?
[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.
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 $\displaystyle (\frac{f}{g})^\prime=\frac{f^\prime g-fg^\prime}{g^2}$.
[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.