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Trigonometry Substitutions


Date: 09/30/97 at 01:26:54
From: Anonymous
Subject: Calc 2 question

I am having problems doing trig substitutions.  Are there any methods  
you can suggest to me? 

I am having problems with the actual substitutions and what to 
substitute for and when to do that substitution. It seems that 
most of my methods lead me in a circle or to unsolvable integrals, 
especially those problems involving the integral of 
[sin^(n)X][cos^(m)X]dx when both n and m are even.  

Your help is greatly appreciated.

Rob Szollosy


Date: 09/30/97 at 14:57:31
From: Doctor Rob
Subject: Re: Calc 2 question

In the particular case you cite, where m and n are both even, you can 
use integration by parts to change the exponents:

  u = sin^(n-1)(X), dv = cos^m(X)*sin(X) dX,
 du = (n-1)*sin^(n-2)(X)*cos(X) dX, v = -cos^(m+1)(X)/(m+1),

for example, will up the exponent of cos(X) by 2 and drop the exponent 
of sin(X) by 2. Alternatively, you can replace sin^2(X) by 1-cos^2(X).
Repeated applications of this can transform your problem into one or 
more integrals of cos^k(X)dX for even values of k. These are done by 
using the identities for cos(k*X) for even k. The smallest ones are:

 cos(2*X) = 2*cos^2(X) - 1
 cos(4*X) = 8*cos^4(X) - 8*cos^2(X) + 1
 cos(6*X) = 32*cos^6(X) - 48*cos^4(X) + 18*cos^2(X) - 1

You can use these to write, for example.

 cos^6(X) = (1/32)*[cos(6*X) + 48*cos^4(X) - 18*cos^2(X) + 1]
          = (1/32)*cos(6*X) + (3/16)*[cos(4*X) + 8*cos^2(X) - 1]
                - (9/16)*cos^2(X) - 1/32
          = (1/32)*cos(6*X) + (3/16)*cos(4*X) + (15/32)*[cos(2*X)+1]
                - 7/32
          = (1/32)*cos(6*X) + (3/16)*cos(4*X) + (15/32)*cos(2*X) - 1/4

This kind of expansion of cos^k(X) in terms of cos(i*X) for 
0 <= i <= K is possible for every k. Now the result is easy to 
integrate. Once done, you can use similar identities for sin(i*X) to 
convert the result back to an expression in sin(X) and cos(X), if you 
so desire.

The tip-off that you need a trigonometric substitution is that you 
have an algebraic expression which involves a quadratic quantity.  
As an example, you know that to integrate Sqrt[1-X^2] dX, you will 
want to use a trig substitution. 

When the quantity has the form a^2 - (b*X+c)^2, then you will be using 
a sine or cosine:  (b*X+c)/a = sin(u), for example, and then the 
quantity will equal a^2*cos^2(u) and dX = (a/b)*cos(u) du. 

When the quantity has the form a^2 + (b*X+c)^2 = (b*X+c)^2 + a^2, you 
will use a tangent or cotangent: (b*X+c)/a = tan(u), for example, and 
then the quantity will equal a^2*sec^2(u) and dX = (a/b)*sec^2(u) du.  

When the quantity has the form (b*X+c)^2 - a^2, you will use a secant 
or cosecant: (b*X+c)/a = sec(u), for example, and then the quantity 
will equal a^2*tan^2(u) and dX = (a/b)*sec(u)*tan(u) du. Notice the 
importance of completing the square in the quadratic quantity.

If you are given a rational function of trigonometric functions to
integrate, such as your example, another way to proceed is to write 
all the trig functions in terms of sin(x) and cos(x), then use the 
universal substitution z = tan(x/2), which gives

  sin(x) = 2*z/(1+z^2), cos(x) = (1-z^2)/(1+z^2), dx = dz/(1+z^2)

This will reduce such integrals to integrals of rational functions 
of z.

Probably there are other useful things that could be said about when 
to use trig substitutions and trig identities in integration problems, 
but this will get you started.

-Doctor Rob,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Trigonometry

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