# Trig' Integrands

Discussion in 'Math' started by amilton542, Jun 8, 2013.

1. ### amilton542 Thread Starter Active Member

Nov 13, 2010
494
64
Hi,

How do you compensate for the difference when computing trig' integrands?

I've computed this breed of integral where there's up to three or four different solutions, where by virtue of which yields a different result by way of the trig' identities.

It sounds easier said than done, but let's say in practice - how do you know what solution is the "correct" one?

2. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
I've never heard of this terminology. I think what you are saying is that any given solution to an integral can be represented in several forms by the application of various identities. Since that is the case there is no "compensation" necessary because the solutions are equivalent. If you can transform one into another by the application of identities then they are the same.

3. ### amilton542 Thread Starter Active Member

Nov 13, 2010
494
64
Ok, lets say you're given an integrand that takes the form $sin^3(x)cos^3(x)$

Now, there are two possible substitutions you can make.

The first case: (Pythagorean Identity)

$F(x) = \int sin^3(x)cos^3(x)\ dx = \int sin(x)[1 - cos^2(x)]cos^3(x)\ dx$

$Let\ u\ = cos(x)\ then\ du = -sin(x)\ dx$

$\Rightarrow F(u) = - \int (1 - u^2)u^3\ du = \int (u^5 - u^3)\ du = \frac{1}{6}u^6 - \frac{1}{4}u^4 + C$

$\Rightarrow F(x) = \frac{1}{6}cos^6(x) - \frac{1}{4}cos^4(x) + C$

The second case: (Double-Angle Substitution)
$F(x) = \int sin^3(x)cos^3(x)\ dx \equiv \frac{1}{8} \int sin^3(2x)\ dx$

$F(x) = \frac{1}{8} \int sin(2x)[1 - cos^2(2x)]\ dx$

$\Rightarrow F(x) = - \frac{1}{16} cos(2x) - \frac{1}{8} \int sin(2x)cos^2(2x)\ dx$

$Let\ u\ = cos(2x)\ then\ du = - 2sin(2x)\ dx$

$\Rightarrow \frac{1}{8} \int sin(2x)cos^2(2x)\ dx$

$= - \frac{1}{16} \int u^2\ du = - \frac{1}{48} u^3 + C = - \frac{1}{48} cos(2x)^3 + C$

$\Rightarrow F(x) = \frac{1}{48}cos^3(2x) - \frac{1}{16}cos(2x) + C$

We take note. When one substitutes an arbitrary value for x (radians), each solution will differ' by some constant because both solutions yield a different result.

With other trig' integrands, I've had up to four solutions on my hands before.

Last edited: Jun 8, 2013
4. ### amilton542 Thread Starter Active Member

Nov 13, 2010
494
64
Look carefully at both solutions.

In the first case:

Divide the index of cosine by 2 and multiply the period by 2. Now 6/2 = 3 and 4/2 = 2.

2^3 = 8 and 2^2 = 4

Now compare this with the second solution.

Multiply the denominator of the first coefficient by 8 and multiply the denominator of the second coefficient by 4 and you arrive at the second solution.

5. ### WBahn Moderator

Mar 31, 2012
18,088
4,917
But each solution includes and arbitrary constant. So if two solutions differ by a constant, they are equivalent because the constant is arbitrary!

More to the point, if you were given initial conditions that allowed you to solve for the arbitrary constant, then each solution would potentially yield a different constant, but now the solutions would no longer differ by a constant because the different constants in each solution would remove the difference.

6. ### WBahn Moderator

Mar 31, 2012
18,088
4,917
I'm not following this at all.

What do you mean by "index of cosine" and "period" here?

You seem to be saying, however, that you can take the first solution and transform it into the second. But I thought your whole point of the thread was that you get solutions that can't be transformed into each other and, hence, are different.

7. ### amilton542 Thread Starter Active Member

Nov 13, 2010
494
64
It was a school boy error folks, ha ha ha.