# Integral Notes

## Derivative and Integral Tables

| $f(x)$ | $\frac{d}{dx}f(x)$ | $\int f(x) dx$ |
|:----------:|:--------------------:|:-----------------------:|
| $x^n$ | $nx^{x-1}$ | $\frac{x^{x+1}}{x+1}+c$ |
| $\frac 1x$ | $\frac{-1}{x^2}$ | $\ln(\lvert x\rvert)+c$ |
| $\sqrt{x}$ | $\frac 1{2\sqrt{x}}$ | $\frac{2\sqrt{x^3}}3+c$ |

### Basic trygonometric functions

| $f(x)$ | $\frac{d}{dx}f(x)$ | $\int f(x) dx$ |
|:----------------:|:------------------:|:-----------------------:|
| $\sin(x)$ | $\cos(x)$ | $-\cos(x)+c$ |
| $\cos(x)$ | $-\sin(x)$ | $\sin(x)+c$ |

### Other trygonometric functions

$\tan(x)=\frac {\sin(x)}{\cos(x)}$
$\cot(x)=\frac 1{\tan(x)}=\frac {cos(x)}{\sin(x)}$
$\sec(x)=\frac 1{\cos(x)}$
$\csc(x)=\frac 1{\sin(x)}$

| $f(x)$ | $\frac{d}{dx}f(x)$ | $\int f(x) dx$ |
|:----------------:|:------------------:|:-----------------------:|
| $\tan(x)$ | $\sec^2(x)$ | $-\log\lvert\cos x\rvert+c$ |
| $\sec^2(x)$ | | $\tan(x)+c$ |
| $\cot(x)$ | $-\csc^2(x)$ | |
| $\csc^2(x)$ | | $\cot(x)+c$ |
| $\sec(x)$ | $\sec(x)\tan(x)$ | |
| $\sec(x)\tan(x)$ | | $\sec(x)+c$ |
| $\csc(x)$ | $-\csc(x)\cot(x)$ | |
| $\csc(x)\cot(x)$ | | $-\csc(x)+c$ |

## Properties of Integrals
1. $\int c\cdot f(x)\,dx = c\cdot\int f(x)\,dx$
2. $\int f(x)\pm g(x)\,dx = \int f(x)\,dx\pm\int g(x)\,dx$

## Integration by Substitution

#### Steps:
1. Pick "u" so the integral is easier.
- usualy it's the inside of something
- the derivative of "u" must be in $\int$ (disregard constants)
2. Transform $\int x\,dx \to \int u\,du$
3. Do integral
4. Translate back to "x"

##### example:
$\int(x^2+1)^{50}\cdot 2x\,dx$
Let:
$u = x^2 + 1$
Then:
$du = 2x\,dx \therefore dx = \frac {du}{2x}$
Now we substitute $u$ and $du$:
$\require{cancel}\int u^{50}\cdot \cancel{2x}\cdot\frac{du}{\cancel{2x}} = \int u^{50}\,du$
So now we can easily solve integral:
$\int u^{50}\,du = \frac {u^{51}}{51} + c$
Then we just need to substitute $x$ back:
$\frac {u^{51}}{51} + c = \frac {(x^2+1)^{51}}{51} + c$
And we're done.
## Integration by Parts

$\int u\,dv = u\,v - \int v\,du$

###### This method is good for solving intgrals with exponential nad trygonometic functions.

##### example:

$\int e^x\sin(x)\,dx$
Let:
$v = e^x \qquad dv = e^x\,dx$
$u = \sin(x) \qquad du = -\cos(x)\,dx$
Then:
$\int e^x\sin(x)\,dx = e^x\sin(x) - \int e^x(-\cos(x))\,dx = e^x\sin(x) + \int e^x\cos(x)\,dx$
Then lets do it again:
$v = e^x \qquad dv = e^x\,dx$
$u = \cos(x) \qquad du = \sin(x)\,dx$
Then:
$\int e^x\sin(x)\,dx = e^x\sin(x) + \int e^x\cos(x)\,dx = e^x\sin(x) + e^x\cos(x) - \int e^x\sin(x)\,dx$
So, we get:
$\int e^x\sin(x)\,dx = e^x\sin(x) + e^x\cos(x) - \int e^x\sin(x)\,dx$
We can add $\int e^x\sin(x)\,dx$ to both sides:
$2*\int e^x\sin(x)\,dx = e^x\sin(x) + e^x\cos(x)$
And divide by $2$, and we get answer:
$\int e^x\sin(x)\,dx = \frac{e^x\sin(x) + e^x\cos(x)}{2}$

## Integration of Rational Functions
$P(x) = a_nx^n + a_{n-1}x^{n-1} +\cdots+a_1x+a_0$
$Q(x) = b_mx^m + b_{m-1}x^{m-1} +\cdots+b_1x+b_0$
$\int\frac {P(x)}{Q(x)}$