Why does my TI-89 give an incorrect result for this integration problem?

[musemonkey]

The integral is

$$I = \int_0^\pi (~3\cos^2(t) - 1~)\sin^2(t)~dt$$.

My TI-89 Titanium says $$I = 0$$, but I know the answer (verified by hand and by mathematica) to be $$-\pi / 8$$. I am 100% sure I entered it correctly into the calculator. What gives?

 

[CompuChip]

If you take pi degrees as the integration boundary, the result is very small. Are you sure you don't have it set to degrees and it's rounding off that small result to zero?
Are you sure you are entering the formula with X instead of T (and T happens to be equal to zero)?

 

[musemonkey]

No I'm in radians. The indefinite integral result that the calculator gives is

$$\int (~3\cos^2(t) - 1~)\sin^2(t)~dt = \frac{\sin(t)\cdot\left(6\cdot\sin^2(t)+1\right)\cdot\cos(t)}{8}-\frac{\mod(2t-\pi,2\pi)}{16}$$,

whereas the correct answer is

$$\int (~3\cos^2(t) - 1~)\sin^2(t)~dt = -\frac{t}{8} + \frac{1}{4}\sin(2t) - \frac{3}{32}\sin(4t)$$.

Differentiating in the calculator the antiderivative that it gives does not give back anything that looks like $$(~3\cos^2(t) - 1~)\sin^2(t)$$.

weird eh?

 

[CompuChip]

Cool, in my day we only had the 83+ which could only do integrations numerically.

Anyway, yep, it's weird. I see some similarities, for example sin(t) cos(t) ~ sin(2t) and mod(2t - pi, 2pi) / 16 would be just t/8 for t > pi/2 (and pi + t/8 for t < pi/2), so some terms seem to agree. Perhaps you can check exactly which terms are there and see what goes wrong... or by looking at the form guess how the TI is doing the integration?

 

[musemonkey]

Well, first, can anyone reproduce this supposed error?

 

[Hurkyl]

Differentiating in the calculator the antiderivative that it gives does not give back anything that looks like $$(~3\cos^2(t) - 1~)\sin^2(t)$$.

So what? Expressions don't have to look the same to be equal...

 

[adriank]

Maybe your TI-89 is broken, or you entered it incorrectly. Mine gives $$-\pi/8$$ for the definite integral and
$$\frac{\sin(t) \cdot \left( 6 (\sin(t))^2 + 1) \cdot \cos(t)}{8} - \frac{t}{8}$$
for the indefinite integral. Differentiating that gives the following mess:
$$\left(\frac{9 \cdot (\sin(t))^2}{4} + 1/8 \right) \cdot (\cos(t))^2 - \frac{3 \cdot (\sin(t))^4}{4} - \frac{(sin(t))^2}{8} - 1/8$$
but subtracting the original integrand from this gives zero. Have you tried that?

Note: my TI-89 (non-Titanium) is HW1, AMS version 2.05, 07/05/2000. What is yours? (If you don't know how to check, at Home, go F1(Tools) - A(About).)

 

[musemonkey]

OK, that's pretty much what I expected. I'll take a photo of the screen and send it to Texas Instruments.

Differentiating the antiderivative and subtracting the original integrand does give zero. The difference between what your and my calculators are producing is that mine gives a modular term instead of -t/8. Cool, so we've narrowed down the problem. Thanks!

 

[CompuChip]

Note that the modular term is precisely equal to -t/8 in one part of the integration region, and it is equal to that up to a constant (which is irrelevant for the differentiation of course) in the rest. So it correctly calculates an anti-derivative (just not the one you'd want, if you expected something continuous ).
However, for the definite integration that will give a factor $(2 \pi) \times (\pi / 2)$ too much.

 

[adriank]

Some more information: a friend of mine also gets $$-\pi/8$$ on his TI-89 Titanium (HW4, AMS 3.10, 07/18/2005).

I'm pretty sure the reason you get zero is because of the discontinuity of the indefinite integral your calculator obtains: what you get is periodic (with period pi), so if you try to obtain the definite integral using that, you will get zero.