integration of cos(x^3)

Apr 2010
13
0
can someone help with this integration?

cos(x^3)

if you can do this?

do you know how to do this?

d/dx (integration of cos(t^3) with range (cube root of x to pi/6)

Please help!!!
 
Mar 2010
107
14
It's not a fundamental integral so you can't integrate it; however, you can use taylor series to get the polynomial of it and then integrate that.
 
Apr 2010
13
0
It's not a fundamental integral so you can't integrate it; however, you can use taylor series to get the polynomial of it and then integrate that.
that's why I don't know how to do....would u mind to teach me how to approach this?
 

Opalg

MHF Hall of Honor
Aug 2007
4,039
2,789
Leeds, UK
can someone help with this integration?

cos(x^3)

if you can do this?

do you know how to do this?

d/dx (integration of cos(t^3) with range (cube root of x to pi/6)

Please help!!!
You don't need to integrate \(\displaystyle \cos(x^3)\) in order to answer this question, which is an exercise in using the Fundamental Theorem of Calculus. Roughly speaking, that theorem says that if you integrate a function and then differentiate it, you get back to where you started from.

Say that \(\displaystyle \int\cos(t^3)\,dt = F(t)\). We don't know what F(t) is, but we do know that \(\displaystyle \tfrac d{dt}F(t) = \cos(t^3)\). Then \(\displaystyle \frac d{dx}\int_{\sqrt[3]x}^{\pi/6}\cos(t^3)\,dt = \frac d{dx}\bigl(F(\pi/6) - F(\sqrt[3]x\,\bigr)\). You can do that differentiation by using the fact that \(\displaystyle \tfrac d{dt}F(t) = \cos(t^3)\), together with the chain rule.
 
Apr 2010
384
153
Canada
You don't need to integrate \(\displaystyle \cos(x^3)\) in order to answer this question, which is an exercise in using the Fundamental Theorem of Calculus. Roughly speaking, that theorem says that if you integrate a function and then differentiate it, you get back to where you started from.

Say that \(\displaystyle \int\cos(t^3)\,dt = F(t)\). We don't know what F(t) is, but we do know that \(\displaystyle \tfrac d{dt}F(t) = \cos(t^3)\). Then \(\displaystyle \frac d{dx}\int_{\sqrt[3]x}^{\pi/6}\cos(t^3)\,dt = \frac d{dx}\bigl(F(\pi/6) - F(\sqrt[3]x\,\bigr)\). You can do that differentiation by using the fact that \(\displaystyle \tfrac d{dt}F(t) = \cos(t^3)\), together with the chain rule.
This is absolutely crucial! (Clapping)
 
Mar 2010
107
14
Oh sorry dude. I didn't see the \(\displaystyle \frac{dy}{dx} \)