•  Integration by substitution.
•  Integration by parts.
•  Integration by trigonometric substitution.
•  Integration of rational functions.

Integration by substitution.

The basic idea with the substitution technique is to simplify the given integral.

Given an integral `int f(x) dx`, come up with a substitution `u = u(x)`.

Find `x = x(u)`  and   `dx = x'(u) du`.

Substitute back into the original function:

`int f(x) dx = int f(x(u))x'(u) du`

If the new integral isn't any easier than the original one, try another substitution.

Once you have found the integral in terms of `u`, you must substitute back to the original variable `x`.

Finding a good substitution isn't always easy but with practice you should get better at it.

Example 1:

Given:  `int x(x^3+4)^5 dx`

Let:     `u=x^3+4`   and    `du=3x dx  ->  x dx=(du)/3`

Then:  `int x(x^3+4)^5 dx = int u^5/3 du = u^6/18+C = (x^3+4)^6/18+C`

Example 2:

Given:  `int sqrt(5x-2) dx`

Let:     `u=5x-2`   and    `du=5 dx ->  dx=(du)/5`

Then:  `int sqrt(5x-2) dx = int u^(1/2)/5 du = 2/15u^(3/2)+C = 2/15(5x-2)^(3/2)+C`

If you are working with definite integrals, it may be possible to express the interval of integration in terms of `u`, in which case you would only have to evaluate the definite integral in terms of `u`. Since the conversion back to `x` is usually quite straight forward, it's just as easy to evaluate the integral over the original `x` interval.

Integration by parts.

Integration by parts can be an effective technique when the integrand is a product of two functions. This method is based on the product rule from derivatives.

                    `(u(x)v(x))' = u'(x)v(x) + u(x)v'(x)`

Since integration is backwards differentiation:

                    `u(x)v(x) = int u'(x)v(x) dx + int u(x)v'(x) dx`

Therefore:   `int u(x)v'(x) dx = u(x)v(x) - int u'(x)v(x) dx`

The trick is to pick `u` and `v` such that the remaining integral will be easier to find than the original integral.

Just remember: `int f(x)g(x) dx ne int f(x)g(x) dx`

The following examples illustrate an effective use of the integration by parts technique:

Example 3:

Given:  `int xcos(x) dx `

Let:     `u=x     and     dv=cos(x) dx`
          `du=dx   and    v=sin(x)`

Since:  `int u(x)v'(x) dx = u(x)v(x) - int u'(x)v(x) dx`

Then:  `int xcos(x) dx = xsin(x)  - int sin(x) dx = xsin(x) + cos(x)+C`

Example 4:

Given:  `int 2xln(x) dx `

Let:     `u=ln(x)     and     dv=2x dx`
          `du=1/x dx   and    v=x^2`

Since:  `int u(x)v'(x) dx = u(x)v(x) - int u'(x)v(x) dx`

Then:  `int 2xln(x) dx = x^2ln(x)  - int x dx = x^2ln(x) - x^2/2+C = x^2(ln(x) - 1/2)+C`

Integration by trigonometric substitution.

This method is particularly effective whenever the integrand involves right triangle math.

Let's examine the three special right triangles shown below:

`x = atan(theta)` `dx = asec^2(theta) d theta` `sqrt(x^2+a^2) = asec(theta)` `theta = tan^(-1)(x/a)`	`x = asin(theta)` `dx = acos(theta) d theta` `sqrt(a^2-x^2) = acos(theta)` `theta = sin^(-1)(x/a)`	`x = asec(theta)` `dx = asec(theta)tan(theta) d theta` `sqrt(x^2-a^2) = atan(theta)` `theta = cos^(-1)(a/x)`

The following trigonometric identities are helpful when making trigonometric substitutions:

`1+tan^2(theta) = sec^2(theta)`

`1-sin^2(theta) = cos^2(theta)`

`sec^2(theta)-1 = tan^2(theta)`

Example 5:

Given:  `int 1/sqrt(16-x^2) dx `

Let:     `x=4sin(theta)                    dx=4cos(theta) d theta                    sqrt(16-x^2)=4cos(theta)`

Then:  `int 1/sqrt(16-x^2) dx = int (4cos(theta) d theta)/(4cos(theta)) = int d theta = theta = sin^(-1)(x/4)+C`

Example 6:

Given:  `int x/sqrt(x^2-4) dx `

Let:     `x=2sec(theta)                    dx=2sec(theta)tan(theta) d theta                    sqrt(x^2-4)=2tan(theta)`

Then:  `int x/sqrt(x^2-4) dx = int (4sec^2(theta)tan(theta) d theta)/(2tan(theta)) = int 2sec^2(theta) d theta = 2tan(theta) = sqrt(x^2-4)+C`

Integration of rational functions.

This method is helpful when you have one polynomial divided by another polynomial, a rational function.

The key to this method is the decomposition of the rational function into partial fractions. In cases where the numerator is of a degree larger than the denominator you must first use long divison or synthetic division to produce an expression plus a rational function. The remaining rational function is then decomposed into partial fractions.

Once the rational function has been simplified, you should find integration of the individual parts easily.

Example 7:

Given:  `int (5x+7)/(x^2+3x+2) dx `

Let:     `(5x+7)/(x^2+3x+2) = A/(x+2)+B/(x+1) => A(x+1)+B(x+2) = 5x+7`

Set:     `x=-2 = A((-2)+1) = 5(-2)+7 => -A = -3 => A = 3`

Set:     `x=-1 = B((-1)+2) = 5(-1)+7 => B = 2`

Then:  `int (5x+7)/(x^2+3x+2) dx = int 3/(x+1)+2/(x+2) dx = 3ln(x+1)+2ln(x+2)+C`

Example 8:

Given:  `int (x^3+5)/(x^2-2x+1) dx `

Since the degree of the numerator is greater than the degree of the denominator, use long division to reduce the integrand:

            `(x^3+5)/(x^2-2x-3) = x+2+(7x+11)/(x^2-2x+1)`

Decompose the fractional part of the reduced integrand into partial fractions:

            `(x^3+5)/(x^2-2x-3) = x+2+(7x+11)/(x^2-2x+1) = x+2+8/(x-3)-1/(x+1)`

Then:  `int (x^3+5)/(x^2-2x+1) dx = int x+2+8/(x-3)-1/(x+1) dx = x^2/2+2x+8ln(x-3)-ln(x+1)+C`

Hopefully, you will find these four techniques helpful when working with more advanced integrands.