Difference between revisions of "User:Able806/integration"

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We then plug in the values of A and B and get,
 
We then plug in the values of A and B and get,
  
−
:<big><math>\int f(x) g'(x)\,dx = f(x) g(x) - \int f'(x) g(x)\,dx,</math></big>
+
:<big><math>\frac{4}{x-3}-\frac{1}{x+2}</math></big>
  
 
Now we can solve the integral.
 
Now we can solve the integral.
Line 56: Line 56:
  
 
===Algebraic Substitution===
 
===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===
 
===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>

Latest revision as of 15:03, April 24, 2008

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:

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

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

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

Therefore we can restate the problem,

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 :,

We let :Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=3} ,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3(3)+11=A(3+2)+B(3-3)}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 20=A(5)+B(0)}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A=4}

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{4}{x-3}-\frac{1}{x+2}}

Now we can solve the integral.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\frac{3x+11}{x^2-x-6}dx=\int\frac{4}{x-3}-\frac{1}{x+2}dx}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\frac{3x+11}{x^2-x-6}dx=\int\frac{4}{x-3}dx-\int\frac{1}{x+2}dx}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\frac{2x}{x^2+3}dx}

We can substitute the term :Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^2+3} with a u. Giving us

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=x^2+3}

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{du}{dx}x^2+3=2x}

We then set the terms equal to du,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du=2xdx}

Now we are ready to rewrite the integral,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\frac{1}{u}du=ln|u|+c}

Now we replace u with the term :Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^2+3} to get,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ln|x^2+3|+c}

We can check this by taking the derivative of :Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ln|x^2+3|} ,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 :Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle sin\theta=\frac{x}{3}} . This allows us to substitute x with :Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3sin\theta} . Now to do the substitution

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 9-x^2=9-(3sin\theta )^2}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 9-(3sin\theta )^2=9-9sin^2\theta }
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 9-9sin^2\theta=9(1- sin^2\theta) }

And by use of trigonometric identities we know that

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1- sin^2\theta=cos^2\theta }
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 9(1- sin^2\theta)=9cos^2\theta }

Therefore

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{9-x^2}=3cos\theta }

We are not done yet, we must also take the derivative of :Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3sin\theta}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dx}{d\theta}3sin\theta=3cos\theta }

By partial derivatives we move the :Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {d\theta}} over.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle dx=3cos\theta d\theta }

Now we are ready to rewrite our integral.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 :Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=3sin\theta} we can see that

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta=sin^{-1}(\frac{x}{3})+c} giving us the final solution.
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\frac{1}{\sqrt{9-x^2}}dx=sin^{-1}(\frac{x}{3})+c }
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