# User:Able806/integration

## 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:

$\int f(x) g'(x)\,dx = f(x) g(x) - \int f'(x) g(x)\,dx,$

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

$\int\frac{3x+11}{x^2-x-6}dx$

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

$\frac{3x+11}{(x-3)(x+2)}=\frac{A}{x-3}+\frac{B}{x+2}$

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

$\frac{3x+11}{(x-3)(x+2)}=\frac{A(x+2)+B(x-3)}{(x-3)(x+2)}$

Therefore we can restate the problem,

$3x+11=A(x+2)+B(x-3)$

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 :$x=-2$,

$3(-2)+11=A(-2+2)+B(-2-3)$
$5=A(0)+B(5)$
$B=-1$

We let :$x=3$,

$3(3)+11=A(3+2)+B(3-3)$
$20=A(5)+B(0)$
$A=4$

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

$\frac{4}{x-3}-\frac{1}{x+2}$

Now we can solve the integral.

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

### 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

$\int\frac{2x}{x^2+3}dx$

We can substitute the term :$x^2+3$ with a u. Giving us

$u=x^2+3$

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

$\frac{du}{dx}x^2+3=2x$

We then set the terms equal to du,

$du=2xdx$

Now we are ready to rewrite the integral,

$\int\frac{2x}{x^2+3}dx=\int\frac{1}{u}du$

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.

$\int\frac{1}{u}du=ln|u|+c$

Now we replace u with the term :$x^2+3$ to get,

$ln|x^2+3|+c$

We can check this by taking the derivative of :$ln|x^2+3|$,

$\frac{d}{dx}ln|x^2+3|=(\frac{1}{x^2+3})(2x)=\frac{2x}{x^2+3}$

### 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

$\int\frac{1}{\sqrt{9-x^2}}dx$

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

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

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

$9-x^2=9-(3sin\theta )^2$
$9-(3sin\theta )^2=9-9sin^2\theta$
$9-9sin^2\theta=9(1- sin^2\theta)$

And by use of trigonometric identities we know that

$1- sin^2\theta=cos^2\theta$
$9(1- sin^2\theta)=9cos^2\theta$

Therefore

$\sqrt{9-x^2}=3cos\theta$

We are not done yet, we must also take the derivative of :$3sin\theta$

$\frac{dx}{d\theta}3sin\theta=3cos\theta$

By partial derivatives we move the :${d\theta}$ over.

$dx=3cos\theta d\theta$

Now we are ready to rewrite our integral.

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

From our trigonometric expression :$x=3sin\theta$ we can see that

$\theta=sin^{-1}(\frac{x}{3})+c$ giving us the final solution.
$\int\frac{1}{\sqrt{9-x^2}}dx=sin^{-1}(\frac{x}{3})+c$