# User:Able806/integration

## Contents

## Methods of Integration

### Integration by Parts

**Integration by parts** is a special technique to facilitate the integration of the product of two functions that otherwise lack an obvious integral. This technique utilizes the insight of the product rule.

The rule for **integration by parts** is stated as follows:

- <math>\int f(x) g'(x)\,dx = f(x) g(x) - \int f'(x) g(x)\,dx,</math>

This rule is often useful when one function is a power of *x* and the other function is a trigonometric function or *e* raised to a power of *x*.

Note that it may be necessary to repeat the **integration by parts** several times, one for each power of *x*.

### Partial Fractions

**Integration by partial fractions** is a technique to facilitate the integration of a rational expression by partial fraction decomposition.

Given an integral

- <math>\int\frac{3x+11}{x^2-x-6}dx</math>

The first step is to factor the denominator as much as possible and get the form of the partial fraction decomposition. Doing this gives,

- <math>\frac{3x+11}{(x-3)(x+2)}=\frac{A}{x-3}+\frac{B}{x+2}</math>

This allows us to split the fraction in to sums by cross multiplying the denominators,

- <math>\frac{3x+11}{(x-3)(x+2)}=\frac{A(x+2)+B(x-3)}{(x-3)(x+2)}</math>

Therefore we can restate the problem,

- <math>3x+11=A(x+2)+B(x-3)</math>

Now we can solve for A and B by subsituting x with a value that allows the term to go to 0. For example,

We let :<math>x=-2</math>,

- <math>3(-2)+11=A(-2+2)+B(-2-3)</math>
- <math>5=A(0)+B(5)</math>
- <math>B=-1</math>

We let :<math>x=3</math>,

- <math>3(3)+11=A(3+2)+B(3-3)</math>
- <math>20=A(5)+B(0)</math>
- <math>A=4</math>

We then plug in the values of A and B and get,

- <math>\frac{4}{x-3}-\frac{1}{x+2}</math>

Now we can solve the integral.

- <math>\int\frac{3x+11}{x^2-x-6}dx=\int\frac{4}{x-3}-\frac{1}{x+2}dx</math>
- <math>\int\frac{3x+11}{x^2-x-6}dx=\int\frac{4}{x-3}dx-\int\frac{1}{x+2}dx</math>
- <math>\int\frac{3x+11}{x^2-x-6}dx=4ln|x-3|-ln|x+2|+c</math>

### Algebraic Substitution

**Integration by Algebraic Substitution** is a technique to facilitate the integration of a rational expression by substituting a more complicated expression with a variable.

Given an integral

- <math>\int\frac{2x}{x^2+3}dx</math>

We can substitute the term :<math>x^2+3</math> with a u. Giving us

- <math>u=x^2+3</math>

We then take the derivative of u with respect to x,

- <math>\frac{du}{dx}x^2+3=2x</math>

We then set the terms equal to du,

- <math>du=2xdx</math>

Now we are ready to rewrite the integral,

- <math>\int\frac{2x}{x^2+3}dx=\int\frac{1}{u}du</math>

We can rewrite the intergal this way due to the subsitution of the x terms with the u terms.

Now we can solve the intergral in terms of u.

- <math>\int\frac{1}{u}du=ln|u|+c</math>

Now we replace u with the term :<math>x^2+3</math> to get,

- <math>ln|x^2+3|+c</math>

We can check this by taking the derivative of :<math>ln|x^2+3|</math>,

- <math>\frac{d}{dx}ln|x^2+3|=(\frac{1}{x^2+3})(2x)=\frac{2x}{x^2+3}</math>

### Trigonometric Substitution

**Integration by Trigonometric Substitution** is a technique to facilitate the integration of a rational expression by substituting a more complicated radical expression with a trigonometric expression.

Given an integral

- <math>\int\frac{1}{\sqrt{9-x^2}}dx</math>

By looking at the radical we can determine that it represents the base of a right triangle by understanding the Pythagorean theorem.

- <math>\sqrt{9-x^2}=3+x</math> where 3 is the hypotenuse and x is the height of the triangle.

This allows us to rewrite the expression to :<math>sin\theta=\frac{x}{3}</math>. This allows us to substitute x with :<math>3sin\theta</math>. Now to do the substitution

- <math>9-x^2=9-(3sin\theta )^2</math>
- <math>9-(3sin\theta )^2=9-9sin^2\theta </math>
- <math>9-9sin^2\theta=9(1- sin^2\theta) </math>

And by use of trigonometric identities we know that

- <math>1- sin^2\theta=cos^2\theta </math>
- <math>9(1- sin^2\theta)=9cos^2\theta </math>

Therefore

- <math>\sqrt{9-x^2}=3cos\theta </math>

We are not done yet, we must also take the derivative of :<math>3sin\theta</math>

- <math>\frac{dx}{d\theta}3sin\theta=3cos\theta </math>

By partial derivatives we move the :<math>{d\theta}</math> over.

- <math>dx=3cos\theta d\theta </math>

Now we are ready to rewrite our integral.

- <math>\int\frac{1}{\sqrt{9-x^2}}dx=\int\frac{3cos\theta}{3cos\theta}d\theta=\int d\theta=\theta+c</math>

From our trigonometric expression :<math>x=3sin\theta</math> we can see that

- <math>\theta=sin^{-1}(\frac{x}{3})+c</math> giving us the final solution.

- <math>\int\frac{1}{\sqrt{9-x^2}}dx=sin^{-1}(\frac{x}{3})+c </math>