[TI-M] Re: An Integral, hehehe

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[TI-M] Re: An Integral, hehehe


> /  5+x^2
> |  ----- dx
> /  9-x^3

Here are some thoughts; everything used here is taught in Calc I and II (or
AP Calc AB).

This looks ripe for a trig substitution, especially after we see an
arctangent in the CAS's answer.  The denominator can be replaced:

9-x^3 = sqrt(9-x^3)^2 = sqrt(3^2-(x^(3/2))^2)^2

Now we can set up a triangle and use the pytagorean theorem to simplify the
sqrt portion (I normally prefer to put the straight side on the right, but
this is likely to format better for people using non-monospaced fonts):

|\
| \
|  \
|   \
|____\


The hypotenuse can be length 3, and the left side can be length x^(3/2), and
we'll call the angle in the lower right theta, or T for short.  By the
pytagorean theorem (leg1^2+leg2^2=hypot^2), we can find the length of that
bottom side.  And since we've set up the triangle conveniently, the length
of that leg works out to:

sqrt(hypot^2-leg^2) = sqrt(3^2-(x^(3/2))^2)

Notice how we've seen that value before.  We know that the ratio of that leg
to the side of constant length (the hypotenuse, length 3) is equal to
adjacent/hypot = cos(T).  Multiply both sides by that hypot, and we get:

sqrt(3^2-(x^(3/2))^2) = 3cos(T)

And the denominator of our original equation is now (3cos(T))^2 = 9cos(T).

We also need to get an equation for x, which is gonna get a bit ugly, but
comes out to:

x = (3sin(T))^(2/3)

So we can replace the x^2 term in the numerator with (3sin(T))^(4/3), and we
also see:

dx = 3^(2/3)*(2/3)*cos(T)*(sin(T))^(-1/3)

Put it all together and you get something too complex to format in an email,
so I'll skip a little bit.  You can pull out the constant multiple, which
after some simplification turns out to be 2*3^(-7/3).  Don't lose that =)

Let's see what it's down to now, without the constant multiple:

/    5-(3sin(T))^(4/3)
|  --------------------- dT
/  cos(T)*(sin(T))^(1/3)

I hope I haven't screwed anything up yet. . . I've done this on paper and
calculator and I'm doing it again as I type it, so I hope not.

That fraction clearly splits into two easily.  The left one is going to look
rally ugly, but on the right one the two sin's will cancel out nicely.  That
becomes 3^(4/3)*tan(T), which integrates to (refer to a trig table here):

-1*3^(4/3)*ln(|cos(T)|)

The left side is where the integral goes horribly wrong.  Anybody else have
any insight on that one?

After you piece it all together you're still going to have trig of theta,
but using the original triangle you can replace and you're back in terms of
x again.

So I tried filling in what I had after all of this.  The resulting equation
isn't identical to the calculator's original answer, although it's very
close.  Specifically, if I solve the equations and try filling in 5 for x,
I'm getting this:

Calc's Answer:   .667187*(arctan(3.35296)+1.72144)
My Answer:       .667187*(arctan(3.35296)-1.93069)

(My answer obtained by integrating what I could of the integral myself, and
leaving that last portion - which was "on the left" - to the calc to
integrate.)

So I'm very close now, but it looks like I made a small mistake somewhere.
If somebody has the patience to retrace my steps, I'm sure it was something
stupid.  But it took long enough to get this far, so I'll leave you with
this and hope someone else feels like finishing it off.

    -Scott
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