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Two Ways to Integrate


Date: 04/15/98 at 12:13:26
From: Lin
Subject: integration

Dear Dr Math,

I've often wondered how you would integrate the following sum:

   Int[sin^4(x)cos^6(x)]dx 

We have recently just done integration by parts and other methods 
of integration, but we never ever did an example like this. I don't 
know how to go about it. Could you please help me? Thank you.

Lin


Date: 04/15/98 at 15:27:32
From: Doctor Rob
Subject: Re: integration

As is often the case with such integrals, there is more than one way 
to proceed. Probably the simplest uses trigonometric identities and 
the sum and difference of angle formulas for sine and cosine.

Recall that:

   sin(A-B) = sin(A)*cos(B) - cos(A)*sin(B)
   sin(A+B) = sin(A)*cos(B) + cos(A)*sin(B)

Then, adding and dividing by 2, you get:

   (1/2)*[sin(A-B)+sin(A+B)] = sin(A)*cos(B)          (Equation 1)

Similarly:

   cos(A-B) = cos(A)*cos(B) + sin(A)*sin(B)
   cos(A+B) = cos(A)*cos(B) - sin(A)*sin(B)

Then, adding and dividing by 2, and subtracting and dividing by 2,
you get:

   (1/2)*[cos(A-B)+cos(A+B)] = cos(A)*cos(B),         (Equation 2)
   (1/2)*[cos(A-B)-cos(A+B)] = sin(A)*sin(B).         (Equation 3)

Now your integrand can be rewritten using these equations to be a sum 
of fractions times first powers of sines or cosines of integer 
multiples of x. Each use reduces the "degree" by 1. Here is how it 
starts:

   sin^4(x)*cos^6(x) = [sin(x)*cos(x)]^4*cos^2(x)
                     = [(1/2)*sin(2*x)]^4*cos^2(x)
                     = (1/16)*sin^4(2*x)*cos^2(x)
                     = (1/16)*sin^2(2*x)*[sin(2*x)*cos(x)]^2
                     = (1/16)*sin^2(2*x)*(1/2)^2*[sin(x)+sin(3*x)]^2
                     = (1/64)*[sin(2*x)*sin(x)]^2 +
                         (1/64)*2*[sin^2(x)]*[sin(x)*sin(3*x)] +
                         (1/64)*[sin(2*x)*sin(3*x)]^2
                     = ...
                     = 3/256 + cos(2*x)/256 - cos(4*x)/64 -
                         3*cos(6*x)/512 + cos(8*x)/256 + cos(10*x)/512

Now this integrand is easy to integrate. If you need to, at the end 
you can convert the trigonometric functions back to ones in terms of 
sin(x) and cos(x).

Another way is to use integration by parts.  Let:

    u = sin^m(x)*cos^(n-1)(x)
   dv = cos(x)*dx
    v = sin(x)
   du = sin^m(x)*(n-1)*cos^(n-2)(x)*(-sin(x)) +
          cos^(n-1)(x)*m*sin^(m-1)(x)*cos(x)]*dx
      = sin^(m-1)(x)*cos^(n-2)(x)*[-(n-1)*sin^2(x)+m*cos^2(x)]*dx

So:

   Integral[u*dv] = u*v - Integral[v*du]

   Integral[sin^m(x)*cos^n(x)*dx]
      = sin^(m+1)(x)*cos^(n-1)(x) +
         (n-1)*Integral[sin^(m+2)(x)*cos^(n-2)(x)*dx] -
          m*Integral[sin^m(x)*cos^n(x)*dx]
      = sin^(m+1)(x)*cos^(n-1)(x) +
         (n-1)*Integral[sin^m(x)*(1-cos^2(x))*cos^(n-2)(x)*dx] -
         m*Integral[sin^m(x)*cos^n(x)*dx],
      = sin^(m+1)(x)*cos^(n-1)(x) +
         (n-1)*Integral[sin^m(x)*cos^(n-2)(x)*dx] -
         (m+n-1)*Integral[sin^m(x)*cos^n(x)*dx] 

Adding (m+n-1)*Integral[sin^m(x)*cos^n(x)*dx to both sides, we get:

   (m+n)*Integral[sin^m(x)*cos^n(x)*dx]
      = sin^(m+1)(x)*cos^(n-1)(x) + 
         (n-1)*Integral[sin^m(x)*cos^(n-2)(x)*dx]

Dividing both sides by (m+n), we get:

   Integral[sin^m(x)*cos^n(x)*dx]
      = (1/[m+n])*sin^(m+1)(x)*cos^(n-1)(x) +
         ([n-1]/[m+n])*Integral[sin^m(x)*cos^(n-2)(x)*dx]

Similarly:

   Integral[sin^m(x)*cos^n(x)*dx]
      = -(1/[m+n])*sin^(m-1)(x)*cos^(n+1)(x) +
         ([m-1]/[m+n])*Integral[sin^(m-2)(x)*cos^n(x)*dx]

You can use the first formula with m = 4 and n = 6, then with
m = n = 4, then with m = 4 and n = 2, and reduce the integral to one 
of integrating sin^4(x)*dx. Then you can use the second formula with 
n = 0 and m = 4, then with n = 0 and m = 2, to reduce the integral to 
one of integrating 1*dx, which you can do (I hope).

Another approach is to use sin^2(x) = 1 - cos^2(x) to write everything
in terms of cosines, then use the first integral formula above with 
m = 0, which simplifies to:

   Integral[cos^n(x)*dx)] = (1/n)*sin(x)*cos^(n-1)(x) +
                              ([n-1]/n)*Integral[cos^(n-2)(x)*dx]

Using this repeatedly reduces the exponent of cos(x) in the integrand 
by 2 each time until you are reduced to integrating 1*dx, again.

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

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