# How Does Substitution Affect Double Integration and Differentiation?

• KungPeng Zhou
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?

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.

anuttarasammyak said:
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

What you posted:
KungPeng Zhou said:
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).

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