The Corporeal Tutorial: Substitution

How do you choose your value for U? (a slight pun on words) I wanted to post a problem that demonstrates the power of choosing the right value for U. Hopefully you will walk away with a better appreciation for U-substitution. SOOOO... lets follow two different approaches to an indefinite integral problem! YAY :-)

Problem:

$\int \left [ \frac{18tan^{2}xsec^{2}x}{(2+tan^{3}x)^{2}} \right ]dx$

(for those die-hard calculus lovers, like myself, I would say stop and calculate before looking ahead!! No peeking!)

Ok, now that the weirdos are gone lets start on the first approach. This approach is cumbersome in comparison to the second.

$u=tanx ; \frac{du}{dx}=sec^{2}x ; du=sec^{2}x(dx)$

$\int \left [ \frac{18u^{2}}{(2+u^{3})^{2}} \right ]du$

$v=u^{3} ; dv=3u^{2}du$

$\int \left [ \frac{6}{(2+v)^{2}} \right ]dv$

$w=2+v ; dw=dv$

$\int \left [ \frac{6}{w^{2}} \right ]dv=-\frac{6}{w}$

$\therefore \int \left [ \frac{18tan^{2}xsex^{2}x}{(2+tan^{3})^{2}} \right ]dx=-\frac{6}{2+tan^{3}x}+C$

Now, you will find this pretty amazing! Lets try with the "other guys" U. (hope you caught the Tommy Boy reference; if not i forgive you) I'm not going to rewrite the problem, just look up, lol.

$u=2+tan^{3}x ; du=(3tan^{2}sec^{2}x)dx$

$6\int \left [ u^{-2} \right ]du=-\frac{6}{2+tan^{3}x}+C$

Point here being that your value of U can make a problem 10X longer or it could make it a one liner. Out of all the various combination's of U you have to find the one that fits the best :-)

The Corporeal Tutorial

Saturday, June 27, 2009

Substitution

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