I am currently taking an integral calculus class and I am having a
hard time trying to solve this extra credit problem. This problem
involves Special Techniques of Integration comparing 3 integrals
involving the integration of a the trig function sine raised to a
power all over the trig function sine raised to the same power plus
the trig function cosine raised to that same power. For instance they
want us to solve and find a common method to integrate these 3
functions.

INTEGRAL( sin[x] / ( sin[x] + cos[x] ) , x , 0 , pi/2)
INTEGRAL( (sin[x])^2 / ( (sin[x])^2 + (cos[x])^2 ) , x , 0 , pi/2)
INTEGRAL( (sin[x])^10 / ( (sin[x])^10 + (cos[x])^10 ) , x , 0 , pi/2)

It gives me a hint as how to go about integrating that function. It
says that the first example is a rational function in sin[x] and
cos[x] and it ususally worked with the substitution u = tan [x/2] and
other trig formulas. Set up the substitution until you get an integral
in u, but do not evaluate. The hardest part is this: it says let f be
a continuous function over the closed interval [0,a]. Then let
	INTEGRAL( f(x) , x , 0 , a) = INT( f(a-x) , x , 0 , a )
Use a u-substitution to verify this. Also how does the graph of f(a-x)
compare to the graph of f(x)? Use this to evaluate :
	INT( (sin[x])^n / ( (sin[x])^n + (cos[x])^n ) , x , 0 , pi/2)
for every n positive integer.

Thank you very much for your assistance.
Felipe
philpre@hotmail.com

Hi, philpre-ga:

There are several parts of this problem.  Let's identify and tackle
them in order.

First you are asked to rewrite the first integral:

INTEGRAL( sin(x)/( sin(x) + cos(x) ), x, 0, pi/2)

using the standard substitution u = tan(x/2) for such problems
(without being required to carry out the integration).

Now:

sin(x) = 2 sin(x/2) cos(x/2)
       = 2 tan(x/2)/sec^2(x/2)
       = 2 u/(1 + u^2)

cos(x) = cos^2(x/2) - sin^2(x/2)
       = (1 - tan^2(x/2))/sec^2(x/2)
       = (1 - u^2)/(1 + u^2)

Also du = (1/2) sec^2(x/2) dx, so:

dx = 2 cos^2(x/2) du
   = 2/(1 + u^2) du

where we have made use of such identities as:

cos(A) = 1/sec(A)
sec^2(A) = 1 + tan^2(A)

Putting these together to rewrite the first integral:

INTEGRAL sin(x) / ( sin(x) + cos(x) ) dx
 = INTEGRAL (2u/(1 + 2u - u^2))(2/(1 + u^2)) du

The strength of this technique is that it applies to the indefinite
integral and hence to respective definite integrals with any limits of
integration.  On the other hand this approach can produce a very
unwieldy looking rational expression in u even for the simplest case
(n = 1) contemplated here.

As jdog-ga has already demonstrated, it is true that:

INTEGRAL f(x) dx OVER [0,a]
 = INTEGRAL f(a-x) dx OVER [0,a]

since the change of variable u = a-x induces two sign changes, which
cancel one another (du = -dx and also a change of sign in swapping
limits of integration).

The geometric interpretation given by jdog-ga of this is correct and
can be expressed compactly as:

Reflect the graph of f(x) about the vertical line x = a/2.

Since the interval of integration [0,a] is symmetric about this
midpoint, the area under the curve remains the same.

As you might suspect from how this extra credit problem is
constructed, the evalution of the indicated family of definite
integrals:

I_n = INTEGRAL( sin^n(x)/(sin^n(x) + cos^n(x)), x , 0 , pi/2)

can be done easily using the above "reflection" idea.

Notice that for the specific interval [0, pi/2]:

sin(pi/2 - x) = cos(x)
cos(pi/2 - x) = sin(x)

so that reflection about x = pi/4 amounts to interchanging the
appearances of the sine and cosine functions.  Therefore:

I_n 
 = INTEGRAL( sin^n(x)/(sin^n(x) + cos^n(x)), x, 0, pi/2)
 = INTEGRAL( cos^n(x)/(cos^n(x) + sin^n(x)), x, 0, pi/2)

Interchanging appearances of sine and cosine left the denominator
fixed.  If we add these two (equal) quantities, their integrands sum
to 1 identically.  So:

2 I_n 
 = INTEGRAL( 1, x, 0, pi/2)
 = pi/2

I_n = pi/4 for n = 1,2,3,...

In this way these definite integral are evaluated for every positive
integer n, and without "sweat" (or tears!).

regards, mathtalk-ga

This was a VERY VERY big help. Thank you for your prompt response. I
will definitely consider using Answers.Google.com any time I have any
problem or question I need answered.

since you think showing that the shift does not affect the value of
the integral is the hardest part, lets see if you can finish off the
problem once you've seen it. We start with the following integral:

INT( f(a-x), x, 0, a )

Let t = a-x. Then we say dt = -dx (equivalenty, dx = -dt), and use the
equation to calculate the new limits of integration (left: a = a-0.
right: 0 = a-a).

INT( -f(t), t, a, 0)

By the linearity of integration, we know we can 'pull out the negative
sign.'

-INT( f(t), t, a, 0)

By switching the limits of integration, we again negate the integral,
so we get

INT( f(a-x), x, 0, a ) = --INT( f(t), t, 0, a ) = INT( f(x), x, 0, a )

(noting that we can change the 'dummy variable' at will...such
nuisances aren't necessary in the rigorous treatment of the subject)

As for the graphs, the graph of f(a-x) is just the reflection across
the y-axis followed by a shift left by a of the graph of f(x).