How Does Substitution Affect Double Integration and Differentiation?

[KungPeng Zhou]

Homework Statement

\frac{d^{2}}{dx^{2}}\int_{0}^{x}(\int_{1}^{sint}\sqrt{1+u^{4}}du)dt

Relevant Equations

FTC1

\frac{d^{2}}{dx^{2}}\int_{0}^{x}(\int_{1}^{sint}\sqrt{1+u^{4}}du)dt=\frac{d}{dx}\int_{0}^{sinx}(\sqrt{1+u^{4}})du
then we let m=sinx，so x=arcsinx，then we get \frac{d}{dx}\int_{0}^{sinx}(\sqrt{1+u^{4}})du=\frac{dm}{dx}\frac{d}{dm}\int_{0}^{m}(\sqrt{1+u^{4}})du=\sqrt{1+m^{4}}\frac{dm}{dx}，then we get the answer easily
Is this method correct?

 

[anuttarasammyak]

I observe your math as below
***********
$$\frac{d^{2}}{dx^{2}}\int_{0}^{x}(\int_{1}^{sint}\sqrt{1+u^{4}}du)dt=\frac{d}{dx}\int_{0}^{sinx}(\sqrt{1+u^{4}})du$$
then we let m=sinx，so x=arcsinx，then we get
$$\frac{d}{dx}\int_{0}^{sinx}(\sqrt{1+u^{4}})du=\frac{dm}{dx}\frac{d}{dm}\int_{0}^{m}(\sqrt{1+u^{4}})du=\sqrt{1+m^{4}}\frac{dm}{dx}$$
，then we get the answer easily
Is this method correct?
************
Is that what you mean ? It seems OK.

 

[KungPeng Zhou]

I observe your math as below
***********
$$\frac{d^{2}}{dx^{2}}\int_{0}^{x}(\int_{1}^{sint}\sqrt{1+u^{4}}du)dt=\frac{d}{dx}\int_{0}^{sinx}(\sqrt{1+u^{4}})du$$
then we let m=sinx，so x=arcsinx，then we get
$$\frac{d}{dx}\int_{0}^{sinx}(\sqrt{1+u^{4}})du=\frac{dm}{dx}\frac{d}{dm}\int_{0}^{m}(\sqrt{1+u^{4}})du=\sqrt{1+m^{4}}\frac{dm}{dx}$$
，then we get the answer easily
Is this method correct?
************
Is that what you mean ? It seems OK.

Yes.Thank you

 

[Mark44]

What you posted:

Homework Statement: \frac{d^{2}}{dx^{2}}\int_{0}^{x}(\int_{1}^{sint}\sqrt{1+u^{4}}du)dt
Relevant Equations: FTC1

\frac{d^{2}}{dx^{2}}\int_{0}^{x}(\int_{1}^{sint}\sqrt{1+u^{4}}du)dt=\frac{d}{dx}\int_{0}^{sinx}(\sqrt{1+u^{4}})du
then we let m=sinx，so x=arcsinx，then we get \frac{d}{dx}\int_{0}^{sinx}(\sqrt{1+u^{4}})du=\frac{dm}{dx}\frac{d}{dm}\int_{0}^{m}(\sqrt{1+u^{4}})du=\sqrt{1+m^{4}}\frac{dm}{dx}，then we get the answer easily
Is this method correct?

Same but using LaTeX:
Homework Statement: $\frac{d^{2}}{dx^{2}}\int_{0}^{x}(\int_{1}^{\sin t}\sqrt{1+u^{4}}du)dt$
Relevant Equations: FTC1

$\frac{d^{2}}{dx^{2}}\int_{0}^{x}(\int_{1}^{\sin t}\sqrt{1+u^{4}}du)dt=\frac{d}{dx}\int_{0}^{\sin x}(\sqrt{1+u^{4}})du$
then we let $m=\sin x$，so $x=arcsin x$，then we get $\frac{d}{dx}\int_{0}^{\sin x}(\sqrt{1+u^{4}})du=\frac{dm}{dx}\frac{d}{dm}\int_{0}^{m} (\sqrt{1+u^{4}})du=\sqrt{1+m^{4}}\frac{dm}{dx}$，then we get the answer easily
Is this method correct?

Your LaTeX is pretty good, but you need to surround each equation with a pair of # characters (inline LaTeX) or a pair of $ characters (standalone LaTeX).