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Integration is the inverse function of [[derivative|derivation]], and is related to it by the [[Fundamental Theorem of Calculus]].
Integration is the inverse function of [[derivative|derivation]], and is related to it by the [[Fundamental Theorem of Calculus]].
==Methods of Integration==
===Integration by Parts===
'''Integration by parts''' is a special [[Techniques of integration|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:
:<big><math>\int f(x) g'(x)\,dx = f(x) g(x) - \int f'(x) g(x)\,dx,</math></big>
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 [[Techniques of integration|technique]] to facilitate the integration of a rational expression by partial fraction decomposition.
Given an integral
:<big><math>\int\frac{3x+11}{x^2-x-6}dx</math></big>
The first step is to factor the denominator as much as possible and get the form of the partial fraction decomposition. Doing this gives,
:<big><math>\frac{3x+11}{(x-3)(x+2)}=\frac{A}{x-3}+\frac{B}{x+2}</math></big>
This allows us to split the fraction in to sums by cross multiplying the denominators,
:<big><math>\frac{3x+11}{(x-3)(x+2)}=\frac{A(x+2)+B(x-3)}{(x-3)(x+2)}</math></big>
Therefore we can restate the problem,
:<big><math>3x+11=A(x+2)+B(x-3)</math></big>
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 :<big><math>x=-2</math></big>,
:<big><math>3(-2)+11=A(-2+2)+B(-2-3)</math></big>
:<big><math>5=A(0)+B(5)</math></big>
:<big><math>B=-1</math></big>
We let :<big><math>x=3</math></big>,
:<big><math>3(3)+11=A(3+2)+B(3-3)</math></big>
:<big><math>20=A(5)+B(0)</math></big>
:<big><math>A=4</math></big>
We then plug in the values of A and B and get,
:<big><math>\frac{4}{x-3}-\frac{1}{x+2}</math></big>
Now we can solve the integral.
:<big><math>\int\frac{3x+11}{x^2-x-6}dx=\int\frac{4}{x-3}-\frac{1}{x+2}dx</math></big>
:<big><math>\int\frac{3x+11}{x^2-x-6}dx=\int\frac{4}{x-3}dx-\int\frac{1}{x+2}dx</math></big>
:<big><math>\int\frac{3x+11}{x^2-x-6}dx=4ln|x-3|-ln|x+2|+c</math></big>
===Algebraic Substitution===
'''Integration by Algebraic Substitution''' is a [[Techniques of integration|technique]] to facilitate the integration of a rational expression by substituting a more complicated expression with a variable.
Given an integral
:<big><math>\int\frac{2x}{x^2+3}dx</math></big>
We can substitute the term :<big><math>x^2+3</math></big> with a u. Giving us
:<big><math>u=x^2+3</math></big>
We then take the derivative of u with respect to x,
:<big><math>\frac{du}{dx}x^2+3=2x</math></big>
We then set the terms equal to du,
:<big><math>du=2xdx</math></big>
Now we are ready to rewrite the integral,
:<big><math>\int\frac{2x}{x^2+3}dx=\int\frac{1}{u}du</math></big>
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.
:<big><math>\int\frac{1}{u}du=ln|u|+c</math></big>
Now we replace u with the term :<big><math>x^2+3</math></big> to get,
:<big><math>ln|x^2+3|+c</math></big>
We can check this by taking the derivative of :<big><math>ln|x^2+3|</math></big>,
:<big><math>\frac{d}{dx}ln|x^2+3|=(\frac{1}{x^2+3})(2x)=\frac{2x}{x^2+3}</math></big>
===Trigonometric Substitution===
'''Integration by Trigonometric Substitution''' is a [[Techniques of integration|technique]] to facilitate the integration of a rational expression by substituting a more complicated radical expression with a trigonometric expression.
Given an integral
:<big><math>\int\frac{1}{\sqrt{9-x^2}}dx</math></big>
By looking at the radical we can determine that it represents the base of a right triangle by understanding the [[Pythagorean theorem]].
:<big><math>\sqrt{9-x^2}=3+x</math></big> where 3 is the hypotenuse and x is the height of the triangle.
This allows us to rewrite the expression to :<big><math>sin\theta=\frac{x}{3}</math></big>. This allows us to substitute x with :<big><math>3sin\theta</math></big>.
Now to do the substitution
:<big><math>9-x^2=9-(3sin\theta )^2</math></big>
:<big><math>9-(3sin\theta )^2=9-9sin^2\theta </math></big>
:<big><math>9-9sin^2\theta=9(1- sin^2\theta) </math></big>
And by use of trigonometric identities we know that
:<big><math>1- sin^2\theta=cos^2\theta </math></big>
:<big><math>9(1- sin^2\theta)=9cos^2\theta </math></big>
Therefore
:<big><math>\sqrt{9-x^2}=3cos\theta </math></big>
We are not done yet, we must also take the derivative of :<big><math>3sin\theta</math></big>
:<big><math>\frac{dx}{d\theta}3sin\theta=3cos\theta </math></big>
By partial derivatives we move the :<big><math>{d\theta}</math></big> over.
:<big><math>dx=3cos\theta d\theta </math></big>
Now we are ready to rewrite our integral.
:<big><math>\int\frac{1}{\sqrt{9-x^2}}dx=\int\frac{3cos\theta}{3cos\theta}d\theta=\int d\theta=\theta+c</math></big>
From our trigonometric expression :<big><math>x=3sin\theta</math></big> we can see that
:<big><math>\theta=sin^{-1}(\frac{x}{3})+c</math></big> giving us the final solution.
:<big><math>\int\frac{1}{\sqrt{9-x^2}}dx=sin^{-1}(\frac{x}{3})+c </math></big>
===External Links===
===External Links===