How to determine whether two complex trig equation is identical.

[PrudensOptimus]

I believe, many of us had this problem:


After finding the integral of some function, we wonder, what the right answer would be. So we goto http://integrals.wolfram.com to look up the answer. The answer the "Integral machine" gave is not always in the same form our answers are.


For example: ∫ Cos[x]^3 = u - u^3/3, where u = sin x.

The answer the integral machine outputs would be something more nastier...


Can someone explain, how do you generally determine whether an expression is identical to another expression(usually in a more complex form)?

[Tom Mattson]

I don't know of a general method, but you can handle trigonometric cases by using the relevant identities. For instance, in the example you cited you have:

∫cos3(x)dx=sin(x)-(1/3)sin3(x)+C

Wolfram's "Integral Machine" gave this answer:

∫cos3(x}dx=(3/4)sin(x)+(1/12)sin(3x)+C

(note, in both cases I added the "+C" in myself).

The obvious difficulty in comparing the above antiderivatives is in the "3x" argument in the second one. You need to use an identity that reduces the argument to "x" in every term.

The identity is:

sin3(x)=(3/4)sin(x)-(1/4)sin(3x)

[PrudensOptimus]

isn't Sin3x also ((1-cos2x)/2)*sinx?

[Tom Mattson]

Originally posted by PrudensOptimus
isn't Sin3x also ((1-cos2x)/2)*sinx?

No, sin2(x)=(1/2)(1-cos(2x)).

[phoenixthoth]

one way that will at least disprove the two formula's equality would be plugging in a few numbers.

i tried to see if the integrals site will accept the command FullSimplify or Simplify in the integrand. first of all, to see if that would even help, i inputted 0 for the integrand. got nothing. so i inputted 1, for which i got x. the idea would have been to have it integrate 1+FullSimplify[G(x)-H(X)] where G and H are the two things you want to see if are equal. if the answer was x, then they're equal.

i had it integrate 1+(Sin[x]^2)-((1-Cos[2x])/2) and it gave x as the answer, so it seems to Simplify or FullSimplify its answers before drawing them.

so to see if H(x)=G(x), have it integrate 1+H(x)-G(x) and if it gives x then H(x)=G(x). having it integrate H(x)-G(x) to get 0 has worked at least once. the problem with doing that is if it thinks H(x)-G(x) is 0 prior to integrating, it won't like it.

it even knew that sin3(x)=(3/4)sin(x)-(1/4)sin(3x) or at least that ∫sin3(x)=∫(3/4)sin(x)-(1/4)sin(3x)