# Methods of integration

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 $\frac{d}{dx} \sin x=?\,$ This article/section deals with mathematical concepts appropriate for a student in late high school or early university.

This article details several methods of integration for advanced high school or early university student, each with an example.

## Estimation

There are various methods of estimating integrals. This is particularly useful if the rule for integrating is unknown or extremely complicated, as in the case of $e^{x^2}$. The primary method of estimation is the Riemann integral.

Another major method, known as Simpson's rule uses the Riemann integral. Instead of using linear estimations, Simpson's rule allows for the use of parabolas or other higher order expressions to estimate integrals.

For a more detailed treatment, see Simpson's rule.

## Integration by Parts

For a more detailed treatment, see 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 can be proven with the product rule.

The rule for integration by parts is stated as follows:

$\int f(x) g'(x)\,dx = f(x) g(x) - \int f'(x) g(x)\,dx,$
or
$\int u\,dv = uv - \int v\,du\,$

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

For a more detailed treatment, see Partial fractions in integration.
Integration by partial fractions is a technique to facilitate the integration of a rational expression by partial fraction decomposition.

Given an integral

$\int \frac {f(x)}{g(x)}dx$

where f(x) and g(x) are both polynomials, integration by partial fractions shows how to separate the problem into multiple integrals before integrating.

## 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 :x2 + 3 with a u. Giving us

u = x2 + 3

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

$\frac{du}{dx}=2x$

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 integral this way due to the subsitution of the x terms with the u terms.

Now we can solve the integral in terms of u.

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

Now we replace u with the term :x2 + 3 to get,

ln | x2 + 3 | + c

We can check this by taking the derivative of :ln | x2 + 3 | ,

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

## 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 :3sinθ. Now to do the substitution

9 − x2 = 9 − (3sinθ)2
9 − (3sinθ)2 = 9 − 9sin2θ
9 − 9sin2θ = 9(1 − sin2θ)

And by use of trigonometric identities we know that

1 − sin2θ = cos2θ
9(1 − sin2θ) = 9cos2θ

Therefore

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

We are not done yet, we must also take the derivative of :3sinθ

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

By partial derivatives we move the :dθ over.

dx = 3cosθdθ

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 = 3sinθ we can see that

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

## Other methods of integration

The primary methods of integration include: