Solving Integrals Using Chain Rule And Fundamental Theorem Of Calculus: A Step-By-Step Guide.

ⅆⅆx(∫x2sin(t4)ⅆt)=

sin(x4) – x to the fourth

Using the chain rule of differentiation and the fundamental theorem of calculus, we can solve this problem step by step.

Let’s start by focusing on the given integral: ∫x^2sin(t^4)ⅆt

We can use u-substitution by letting u = t^4, which gives us:

du/dt = 4t^3
dt = 1/(4t^3) du

Substituting these into the integral:

∫x^2sin(t^4)ⅆt = ∫sin(u)/(4t^3) * dx^2

Now, we can simplify this expression using the chain rule of differentiation:

ⅆⅆx(∫x^2sin(t^4)ⅆt) = ⅆ/ⅆx ( ∫sin(u)/(4t^3) * x^2)ⅆt

= ∫sin(u)/(4t^3) * d/dx(x^2)ⅆt

Applying the chain rule of differentiation:

= ∫sin(u)/(4t^3) * 2xⅆt

= ∫sin(t^4)/(2t)ⅆt

This integral is not easy to solve directly, so we need to use another method such as integration by substitution. Let’s let w = t^2:

dw/dt = 2t
dt = 1/(2t)dw

Plugging these into the integral:

= (1/2) ∫sin(w^2)ⅆw

This integral cannot be solved analytically in terms of elementary functions, so it must be represented in terms of special functions like the Fresnel integral or the Dawson function.

So, the final answer in terms of special functions is:

ⅆⅆx ( ∫x^2sin(t^4)ⅆt) = (1/2) ∫sin(w^2)ⅆw, where w = t^2.

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