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27 February 2004

 

Integration by Substitution

George E. Hrabovsky, President, MAST

 

What We Did Last Time

We explored the general idea of the chain rule.

Where We Will Go This Time

I will present a useful technique for integration.

Integrating Complicated Functions

Last time we explored the idea of the chain rule for derivatives. This time we will examine a similar technique for integration. Let us say we have a function,

f(x) = sin (x^3 + 2 x^2 - 4 x),

and we want to find its integral.  

F(x) = ?sin (x '^3 + 2 x '^2 - 4 x ') ?x ',

Instead of beginning to crank our way through the enormously complicated expression, we could instead define a new variable,

y = x^3 + 2 x^2 - 4 x .

Our new function becomes,

f(y) = sin y .

Wait a minute, though.  What happens to the differential in the expression for the integral?  We need to relate this to a new differential for y.  Since the expression for y is known, we can take its derivative,

?y/?x = ?/?x (x^3 + 2 x^2 - 4 x)

FormBox[RowBox[{         , RowBox[{=, RowBox[{3 x^2, , +, , 4 x, , -, , 4.}]}]}], TraditionalForm]

We can separate the variables,

?y = (3 x^2 + 4 x - 4) ?x

and then we can solve for ?x,

?x = ?y/(3 x^2 + 4 x - 4) .

So, the integral now looks like,

F(y) = ?sin y ' ?y '/(3 x^2 + 4 x - 4) .

Since the values of x are constants with regards to y, and we know that the integral of the product of a variable and a constant looks like this,

?c a ?a = c?a ?a,

we can rewrite our expression,

F(y) = ?sin y ' ?y '/(3 x^2 + 4 x - 4)

            = 1/(3 x^2 + 4 x - 4) ?sin y ' ?y ' .

            = (-cos y)/(3 x^2 + 4 x - 4)

            = -(cos y)/(3 x^2 + 4 x - 4) .

We must now remove the variable y,

F(y) = -(cos (x^3 + 2 x^2 - 4 x))/(3 x^2 + 4 x - 4) .

Mathematics Challenge from Last Time

The problem was to develop the idea of the chain rule to partial derivatives.  Let us assume that we have a function

a = x y + tan y

and x and y are themselves functions dependent upon another variable t,

x = sin t

and

y = cos t .

Now let's say that we want to find the derivative of a with respect to t.  We might think that we can consider the partial derivatives along with the chain rules,

?a/?t = ?a/?x?x/?t

and

?a/?t = ?a/?y?y/?t .

But wait a minute, we do not need to consider the derivatives of x or y with respect to t as partial derivatives, since they are functions of one variable.  So we then have,

?a/?t = ?a/?x?x/?t

and

?a/?t = ?a/?y?y/?t .

We can add the right-hand sides together to get,

?a/?t = ?a/?x?x/?t + ?a/?y?y/?t .

For our example, this becomes,

?a/?t = y cos t + (x + sec^2 y) (-sin t)

          = cos t cos t +[sin t + sec^2 (cos t)] (-sin t)

          = cos^2 t +[-sin^2 t - sin t sec^2 (cos t)]

          = cos^2 t - sin^2 t - sin t sec^2 (cos t) .

Since,

cos^2 t - sin^2 t   = cos 2 t,

we have

?a/?t = cos 2 t - sin t sec^2 (cos t) .

Another problem we can address is when we have a function, say

a = x y + tan y

and both xFormBox[Cell[TextData[]], TraditionalForm]and y are functions of two other variables,

x = s sin t

and

y = s cos t .

In this case we can take the partial derivative of a with respect to s,

?a/?s = ?a/?x?x/?s + ?a/?y?y/?s

         = y sin t + x cos t .

By substitution we have,

?a/?s = s sin^2 t + s sin t cos t .

By applying standard trigonometric identities we end up with,

?a/?s = s/2 - s/2cos 2 t + s/2sin 2 t .

Similarly,

?a/?t = ?a/?x?x/?t + ?a/?y?y/?t

         = y  s cos t + (x + sec^2 y) cos t .

Again, we can make a substitution,

?a/?t = s^2 cos^2 t   + s sin t cos t + cos t sec^2 (s cos t) .

By applying trigonometric identities we have,

?a/?t = s^2/2 + s^2/2cos 2 t   + s/2 sin 2 t + cos t sec^2 (s cos t) .

Mathematics Challenge

How do higher order partial derivatives work?

Sources About Differential Equations

On-Line:

The ultimate resource for those interested in research related to differential equations is the Electronic Journal of Differential Equations, which is free:

http://ejde.math.swt.edu/

Another nice site that covers lots of topics, though with only a few examples, is the SOS Mathematics page for differential equations:

http://www.sosmath.com/diffeq/diffeq.html

Another free online journal is the Electronic Journal of Qualitative Theory of Differential Equations:

http://www.math.u-szeged.hu/ejqtde/

An online course in differential equations is given at this site:

http://math.stcc.edu/DiffEq/DiffEq.html

Books:

Martin Braun (1993), Differential Equations and Their Applications, Springer-Verlag.  This is a nice book crammed full of applications.

George F. Simmons (1991), Differential Equations with Applications and Historical Notes, McGraw-Hill. This is one of my favorite books about differential equations. It begins at an elementary level and includes nonlinear equations and partial differential equations. My columns will loosely follow this book up to a point and will then take a more geometrical viewpoint.

John W. Dettman (1974), Introduction to Linear Algebra and Differential Equations, McGraw-Hill (reprinted in 1986 by Dover Publications). This is a wonderful book that is also inexpensive. It begins with a discussion of complex numbers and then covers linear algebra and differential equations.

Morris Tenenbaum and Harry Pollard (1963), Ordinary Differential Equations, Harper and Row (reprinted in 1985 by Dover Publications). This is an inexpensive and encyclopaedic treatment of the subject.

Vladimir I. Arnold (1973), Ordinary Differential Equations, MIT Press (as of 1991 in its 8th printing, translated from the Russian by Richard A. Silverman).  This is my favorite book about differential equations at a failry elementary level.  What makes this book so remarkable is its use of geometric methods.  My columns will follow this book after a firm foundation has been laid with traditional methods.

Created by Mathematica  (February 26, 2004).