# integration of cos(x^3)

#### gklove56

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)

#### lilaziz1

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.

#### gklove56

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
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)

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 $$\displaystyle \frac d{dx}\int_{\sqrtx}^{\pi/6}\cos(t^3)\,dt = \frac d{dx}\bigl(F(\pi/6) - F(\sqrtx\,\bigr)$$. You can do that differentiation by using the fact that $$\displaystyle \tfrac d{dt}F(t) = \cos(t^3)$$, together with the chain rule.

• HallsofIvy and AllanCuz

#### AllanCuz

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 $$\displaystyle \frac d{dx}\int_{\sqrtx}^{\pi/6}\cos(t^3)\,dt = \frac d{dx}\bigl(F(\pi/6) - F(\sqrtx\,\bigr)$$. 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)

#### lilaziz1

Oh sorry dude. I didn't see the $$\displaystyle \frac{dy}{dx}$$