Problem: Solving differential equation
Then by chain rule,
Rewrite (1) as
Integrate (3) with respect to gives
where since .
Square (4) gives
Solving for from (5), we obtain
Integrate (6) with respect to yields
since is an even function.
By definition (see “Deriving Two Inverse Functions“),
From “Integration of Trigonometric Expressions ” :
Without using a CAS,
Exercise-1 What is the derivative of hint:
The general form of Leibniz’s Integral Rule with variable limits states:
Suppose satisfies the condition stated previously for the basic form of Leibniz’s Rule (LR-1, see “A Semi-Rigorous Derivation of Leibniz’s Rule“) . In addition, are defined and have continuous derivatives for Then for
(1) can be derived as a consequence of LR-1, the Multivariable Chain Rule, and the Fundamental Theorem of Calculus (FTC):
on the left side of (1) is a function of .
(2) can be expressed as
Hence, by the chain rule,
It follows that
Leibniz’s rule (LR-1) states:
Let be continuous and have a continuous derivative in a domain of plane that includes the rectangle
I will derive LR-1 semi-rigorously as follows:
Integrate (1-1) with respect to from a constant to a variable , we have
While and are functions of , is a constant.
differentiate (1-2) with respect to gives
In the three-dimensional -space, the double integral of a continuous function with two independent variables, , may be interpreted as a volume between the surface and the -plane:
We see from Fig. 1 that on one hand,
but on the other hand,
Since (2-1) and (2-2) amounts to the same thing, it must be true that
In other words, the order of integration can be interchanged.