# [SOLVED] Definite Intgral Involving lnx

#### myngo9191

This is the equation:

$$\displaystyle \frac{d}{dx} \int_{-1}^{lnx^2} (4tsint)^\frac{1}{3} dt$$

and this is what I have:

$$\displaystyle (4xsinx)^\frac{1}{3} \Bigg|_{-1}^{lnx^2}$$

Now where do I go from here?

#### AllanCuz

This is the equation:

$$\displaystyle \frac{d}{dx} \int_{-1}^{lnx^2} (4tsint)^\frac{1}{3} dt$$

and this is what I have:

$$\displaystyle (4xsinx)^\frac{1}{3} \Bigg|_{-1}^{lnx^2}$$

Now where do I go from here?
This follows from the fundemental theorem of calculus and what have there is not correct.

$$\displaystyle \frac{d}{dx} \int_{a}^{g(x)} f(t)dt = f[g(x)]g(x)$$

Can you compute knowing the above?

#### myngo9191

This follows from the fundemental theorem of calculus and what have there is not correct.

$$\displaystyle \frac{d}{dx} \int_{a}^{g(x)} f(t)dt = f[g(x)]g(x)$$

Can you compute knowing the above?
Is this what I should have?

$$\displaystyle (4lnx^2sinlnx^2)^\frac{1}{3} \frac{2x}{x^2} \Bigg|_{-1}^{lnx^2}$$

#### AllanCuz

Is this what I should have?

$$\displaystyle (4lnx^2sinlnx^2)^\frac{1}{3} \frac{2x}{x^2} \Bigg|_{-1}^{lnx^2}$$
From above,

$$\displaystyle \frac{d}{dx} \int_{-1}^{lnx^2} (4tsint)^\frac{1}{3} dt = [4(lnx^2)sin(lnx^2)]^\frac{1}{3} * \frac{2}{x}$$

We do not evaluate here which is what you have in your post. We are finished, this is the answer!

• myngo9191

#### myngo9191

From above,

$$\displaystyle \frac{d}{dx} \int_{-1}^{lnx^2} (4tsint)^\frac{1}{3} dt = [4(lnx^2)sin(lnx^2)]^\frac{1}{3} * \frac{2}{x}$$

We do not evaluate here which is what you have in your post. We are finished, this is the answer!

0o0h!!! Thank you s0o0 much!!! (Yes)