[TI-M] AW: Re: An Integral, hehehe
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[TI-M] AW: Re: An Integral, hehehe
-----Urspr=FCngliche Nachricht-----
Von: ti-math-bounce@lists.ticalc.org
[mailto:ti-math-bounce@lists.ticalc.org] Im Auftrag von Scott Noveck
Gesendet: Donnerstag, 2. Mai 2002 06:27
An: ti-math@lists.ticalc.org
Betreff: [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 =3D sqrt(9-x^3)^2 =3D 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=3Dhypot^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) =3D 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 =3D cos(T). Multiply both sides by that hypot, and we =
get:
sqrt(3^2-(x^(3/2))^2) =3D 3cos(T)
And the denominator of our original equation is now (3cos(T))^2 =3D
9cos(T).
Ist this true: (3cos(T))^2 =3D 9cos(T) ?
Right now I only had time to read through it very fast, but that seems
to be (3cos(T))^2 =3D 9cos(T)^2?!
Christoph
References: