Modified Tabular Integration

Several weeks ago Dr. Qibo Jing an AP teachers at Rancho Solano Preparatory School in Scottsdale, Arizona, posted a new way to approach tabular integration to the AP Calculus Community discussion group.  He calls this the Modified Tabular Method. The algorithm makes repeated integration by parts quicker and more streamlined than the usual method. The usual method is explained here.

Here is Dr. Jing’s method outlined and illustrated with this example \displaystyle I=\int{\left( {{x}^{3}}+7{{x}^{2}} \right)\cos \left( x \right)dx}

Step 1: Identify u and dv in the usual way and rewrite the given integral in terms of u={{x}^{3}}+7{{x}^{2}} and dv=\cos \left( x \right)=\tfrac{d}{dx}\sin \left( x \right).

\displaystyle I=\int{\left( {{x}^{3}}+7{{x}^{2}} \right)\cos \left( x \right)dx}=\int{\left( {{x}^{3}}+7{{x}^{2}} \right)\left( \tfrac{d}{dx}\left( \sin x \right) \right)}

Step 2: The first term of the antiderivative is the product of the two functions that appear in the revised integral:

\displaystyle I=\int{\left( {{x}^{3}}+7{{x}^{2}} \right)\left( \tfrac{d}{dx}\left( \sin x \right) \right)}=\left( {{x}^{3}}+7{{x}^{2}} \right)\sin \left( x \right)\cdots

Step 3: The remaining terms alternate sign. The next term has subtraction sign etc. The first factor of the next term is the derivative of the first factor of the current term. The second factor is the antiderivative of the second factor of the current term – differentiate the first factor; integrate the second factor.

\displaystyle I =\int{\left( {{x}^{3}}+7{{x}^{2}} \right)\left( \tfrac{d}{dx}\left( \sin x \right) \right)}=\left( {{x}^{3}}+7{{x}^{2}} \right)\sin \left( x \right)-\left( 3{{x}^{2}}+14x \right)\left( -\cos \left( x \right) \right)\cdots

Step 4: Alternate signs and repeat step 3 until the term becomes zero or until the original form appears (See next example). DO NOT simplify either factor until you are done as this may change later terms.

Modified A

The simplified result is

\displaystyle I=\left( {{x}^{3}}+7{{x}^{2}}-6x-14 \right)\sin \left( x \right)+\left( 3{{x}^{2}}+14x-6 \right)\cos \left( x \right)+C

Here is a second example.  In this example the original integral returns as the third term.

\displaystyle I=\int{{{e}^{x}}\sin \left( x \right)dx=\int{\sin \left( x \right)\left( \tfrac{d}{dx}{{e}^{x}}dx \right)}}

=\sin \left( x \right)\left( {{e}^{x}} \right)-\cos \left( x \right)\left( {{e}^{x}} \right)+\left( -\sin \left( x \right) \right)\left( {{e}^{x}} \right)

={{e}^{x}}\sin \left( x \right)-\cos \left( x \right)\left( {{e}^{x}} \right)-I

2I={{e}^{x}}\sin \left( x \right)-\cos \left( x \right)\left( {{e}^{x}} \right)

I=\tfrac{1}{2}\left( {{e}^{x}}\sin \left( x \right)-\cos \left( x \right)\left( {{e}^{x}} \right) \right)+C

This really is a simpler algorithm than the usual tabular method.

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