Solving Abs(Cos(x)) Integral Homework

[dobry_den]

Homework Statement

\int \left|\cos t\right| \ dt

The Attempt at a Solution

I divided the interval (-infinity, +infinity) into 2 types of subintervals where cosine is positive and negative respectively. But I'm not sure how to combine these two integrals to get one formula for the whole interval of real numbers. My textbook gives the result as

\sin \left(x - \pi \left(\frac{x}{\pi} + \frac{1}{2} \right)\right) + 2\left(\frac{x}{\pi} + \frac{1}{2} \right)

Do you have any idea how to arrive at such a formula? Thanks a lot in advance!

 

[gamesguru]

That answer can be greatly simplified. It simplifies to \frac{2x}{\pi}. Try it on your calculator.

 

[sutupidmath]

Homework Statement

\int \left|\cos t\right| \ dt

!

Is this supposed to be a definite or indefinite integral? Because if it is just indefinite then i think you missed something, because the answer you provided is defenitely wrong. Since if you take the derivative of your answer you defenitely won't end up with the integrand.
YOu only need to determine where cos(t) changes it's sign, and then integrate separately on those intervals, and you should end up with two answers: -sin(t)+c, on some interval where cos(t) is positive, and sin(t)+c,on the interval where cos(t) is negative.

 

[Schrodinger's Dog]

Is this supposed to be a definite or indefinite integral?

That's what I thought because

\int_{-\infty}^{\infty}\mid\cos(t)\mid=?

Does not converge so the answer their would be some sort of Taylor series I would imagine.

I don't understand the question?

And the general solution is obvious as said above.

 

[HallsofIvy]

That answer can be greatly simplified. It simplifies to \frac{2x}{\pi}. Try it on your calculator.

No, it does not. You've missed a parenthesis.

 

[Schrodinger's Dog]

No, it does not. You've missed a perenthesis.

Oh yeah, I was going to say that as well it actually simplifies to

\sin\left [\frac{(-\pi^2+4x+2\pi)}{2\pi}\right ]

I'd try doing it by hand myself.

But since I don't understand the question anyway seemed a bit pointless?

 

[sutupidmath]

No, it does not. You've missed a parenthesis.

Well, if someone missed a parenthesis, i would say that the OP missed it.

 

[Schrodinger's Dog]

Well, if someone missed a parenthesis, i would say that the OP missed it.

Indeed. It's either that or \sin(\frac{2x}{\pi}+1) though.

 

[dobry_den]

Is this supposed to be a definite or indefinite integral? Because if it is just indefinite then i think you missed something, because the answer you provided is defenitely wrong. Since if you take the derivative of your answer you defenitely won't end up with the integrand.

it's an indefinite integral

 

[sutupidmath]

it's an indefinite integral

Well, i would again say that, that answer as it is given is not correct. But i might be missing something and those answers could be equivalent, i just can't see at first sight that those two answers actually are equivalent.

 

[dobry_den]

the same with me, that's why I'm confused about it.. but still, thanks a lot!

 

[Schrodinger's Dog]

the same with me, that's why I'm confused about it.. but still, thanks a lot!

Must be a misprint then. Because no combination of playing around with brackets produces the actual result of \int \mid\cos(t)\mid\;dt.

 

[gamesguru]

No, it does not. You've missed a parenthesis.

Am I missing something? Because as far as I can see,
<br /> \sin \left(x - \pi \left(\frac{x}{\pi} + \frac{1}{2} \right)\right) + 2\left(\frac{x}{\pi} + \frac{1}{2} \right)=\frac{2x}{\pi}<br />

 

[sutupidmath]

Am I missing something? Because as far as I can see,
<br /> \sin \left(x - \pi \left(\frac{x}{\pi} + \frac{1}{2} \right)\right) + 2\left(\frac{x}{\pi} + \frac{1}{2} \right)=\frac{2x}{\pi}<br />

No you are not missing anything, from what the OP originally stated, Halls thought though that the OP misplaced parenthesees.

Halls might have thought that the expression reads like this:

<br /> \sin \left(x - \pi \left(\frac{x}{\pi} + \frac{1}{2} \right) + 2\left(\frac{x}{\pi} + \frac{1}{2} \right)\right)<br />

which is not what the op indeed wrote.

 

[gamesguru]

No you are not missing anything, from what the OP originally stated, Halls thought though that the OP misplaced parenthesees.

Halls might have thought that the expression reads like this:

<br /> \sin \left(x - \pi \left(\frac{x}{\pi} + \frac{1}{2} \right) + 2\left(\frac{x}{\pi} + \frac{1}{2} \right)\right)<br />

which is not what the op indeed wrote.

Ok, thanks for the clarification. And \frac{2x}{\pi} actually works for any multiple of \pi, but that's not good enough to be the real integral.

 

[Schrodinger's Dog]

No you are not missing anything, from what the OP originally stated, Halls thought though that the OP misplaced parenthesees.

Halls might have thought that the expression reads like this:

<br /> \sin \left(x - \pi \left(\frac{x}{\pi} + \frac{1}{2} \right) + 2\left(\frac{x}{\pi} + \frac{1}{2} \right)\right)<br />

which is not what the op indeed wrote.

That's my fault, because that equation does not match the integral, I assumed there might of been a mistake. Blame me not HOI. However it's still correct that neither equation yields the correct result. Thus the first post from stupidmath is correct, the answer is the absolute value when cos(t) is positive or negative on the intervals given there, not the textbook answer, which might be a misprint?

 

[sutupidmath]

Yeah, defenitely, there seems to be no reasonable way that \int|cos(t)|dt=\sin \left(x - \pi \left(\frac{x}{\pi} + \frac{1}{2} \right)\right) + 2\left(\frac{x}{\pi} + \frac{1}{2} \right) Or the other one, when we change the place of parenteses.

I would say there might been a misplacement of results at the back of the book, because it is not correct this way. End of story!