Goes deeper

Priciple

With goes down to those hard problems, I want to let you know a priciple that we use in general math problem:

We always split hard problem to multiple easy problems or step by step simplify hard problem.

In computer science, we named it Divide & Conquer (D&C algorithms)

Simple one

$$\begin{align*} &\int \frac{1}{a^2 + x^2} dx \\ \\ =& \frac{1}{a^2} \int \frac{1}{1 + (\frac{x}{a})^2} dx &\text{//divide } a^2 \text{in denominator while keep equation balance} \\ \\ =& \frac{1}{a} \int \frac{1}{1 + (\frac{x}{a})^2} d(\frac{x}{a}) \\ \\ =& \frac{1}{a} \cdot \arctan{\frac{x}{a}} + C &\text{//}(\arctan{x})^\prime = \frac{1}{1 + x^2} \end{align*}$$

$$\begin{align*} \\ \\ \\ &\int \frac{1}{\sqrt{a^2 - x^2}} dx ,(a > 0) \\ \\ =& \frac{1}{a} \int \frac{1}{\sqrt{1-(\frac{x}{a})^2}} dx \\ \\ =& 1 \int \frac{1}{\sqrt{1-(\frac{x}{a})^2}} d(\frac{x}{a}) &\text{//divide } a \text{ in x of dx while keep equation balance} \\ \\ =& \arcsin(\frac{x}{a}) + C \\ \\ \\ \end{align*}$$

Let's try to analyze this equation:

$$\frac{1}{a^2} \int \frac{1}{1 + (\frac{x}{a})^2} dx = \frac{a}{a^2} \int \frac{1}{1 + (\frac{x}{a})^2} d(\frac{x}{a})$$

We knew that $x$ of $dx$ is the original function, so do $\frac{1}{a^2}$. Because two of them in a same original function, when $x$ becomes $\frac{x}{a}$, $\frac{1}{a^2}$ have to times $a$.

As for $\int \frac{1}{1 + (\frac{x}{a})^2}$, it must interact with $dx$, then it becomes original function.

You can see $\int (...) d$ as a whole thing. you also can extract a constant from it, we will talk about it later.

$$\begin{align*} \\ \\ \\ &\int \frac{1}{x^2 - a^2} dx ,(a \neq 0) \\ \\ =& \int \frac{1}{(x+a)(x-a)} dx \\ \\ =& \frac{1}{2a} \int (\frac{1}{x-a} - \frac{1}{x+a}) dx \\ \\ \text{//}& \frac{1}{x-a} - \frac{1}{x+a} = \frac{x+a-x+a}{(x-a)(x+a)} = \frac{2a}{(x-a)(x+a)} \\ \\ \text{//}&\text{ Because }2a \neq 1 \text{, divide 2a, that is, times} \frac{1}{2a} \text{in the beginning} \\ \\ =& \frac{1}{2a}(\int \frac{1}{x-a} dx - \int \frac{1}{x+a} dx) \\ \\ =& \frac{1}{2a}(\int \frac{1}{x-a} d(x-a) - \int \frac{1}{x+a} d(x-a)) &\text{//don't forget the basic temple for solving integral eqution} \\ \\ =& \frac{1}{2a} \ln{|\frac{x-a}{x+a}|} + C \\ \\ \\ \end{align*}$$

Complex one

$$\begin{align*} &\int \frac{1}{1 - \sqrt{x}} dx \\ \\ & \text{make } \sqrt{x}=t \text{ , then } x=t^2 \\ \\ =& \int \frac{1}{1-t} d(t^2) \\ \\ =& \int \frac{2t}{1-t} d(t) \\ \\ =& 2 \int \frac{t}{1-t} d(t) \\ \\ =& -2 \int \frac{t}{t-1} d(t) \\ \\ =& -2 \int \frac{t-1+1}{t-1} d(t) \\ \\ =& -2 \int \frac{t-1}{t-1} dt -2 \int \frac{1}{t-1} dt \\ \\ =& -2 \int 1 dt -2 \int \frac{1}{t-1} dt \\ \\ =& -2t - 2 \ln{|t-1|} + C \\ \\ =& -2\sqrt{x} - 2 \ln{|\sqrt{x}-1|} + C \end{align*}$$

Conclusion

1. Keep equation balance

2. Best way to deal with $\sqrt{}$ is to replace it with a symbol

3. Molecular is a constant is more convenient for us to solve equation, especially when it is $1$ (for example: extract a constant out, and put it in front of $\int$)