Integration by factors

I tried to prove this but I fall into a loop when I try to apply integration by factors, that is I prove that the integral is equal to itself.

Any helpfull tips?

Im on my phone, so I cannot see anything past the f². Sine and cosine derivatives are related. You'll likely need that. Maybe use product rule to expand (f * sin)' first

Is it true? Try ##f(x) = \cos x##.

PeroK said:

Is it true? Try ##f(x) = \cos x##.

I forgot to say that ##f(x)## is positive in ##[0,2\pi]##. And of course ##f## is differentiable.

@PeroK let me ask you the dual problem , it holds that $$\int_0^{2\pi} (f(x)\cos x)(f(x)\cos x)'=0$$ right?

But yeah i am not 100% sure if the original problem is true or not .

Delta2 said:

I forgot to say that ##f(x)## is positive in ##[0,2\pi]##. And of course ##f## is differentiable.

Try ##f(x) = \cos^2 x##.

PeroK said:

Try ##f(x) = \cos^2 x##.

Don't see any light in the tunnel, I tried expanding the derivative but it goes a mess, hold on while I try better.

About the other integral you agree it is zero?

Delta2 said:

Don't see any light in the tunnel, I tried expanding the derivative but it goes a mess, hold on while I try better.

About the other integral you agree it is zero?

I think you can't prove it because it isn't true!

PeroK said:

I think you can't prove it because it isn't true!

Ok, the other integral is not equal to zero either?

Where it goes wrong for the other I set $$g(x)=f(x)\cos x$$ so it goes $$\int_0^{2\pi}g(x)g'(x)dx=\frac{1}{2}[g^2(2\pi)-g^2(0)]=0$$ since ##g(2\pi)=g(0)## from the periodicity of f(x) and cosx.

Forgot a factor of ##\frac{1}{2}## in the initial problem, I ll try to edit it.

Delta2 said:

Ok, the other integral is not equal to zero either?

Plug in ##f(x) = \cos^2 x## and see whether the equality holds.

PeroK said:

Plug in ##f(x) = \cos^2 x## and see whether the equality holds.

yes I tried this and by expanding the derivative and with the help of wolfram it gives zero indeed.

I tried the other too (initial problem) and wolfram says that it is true! (for ##f(x)=cos^2(x)## (with the 1/2 factor correction). Beware I am not asking for the indefinite integrals but for the definite integrals within the period interval ##[0,2\pi]## .

Tried ##f(x)=\sin^2 x## and it works again wolfram says the integrals are equal to ##3\frac{\pi}{8}##.

Delta2 said:

Tried ##f(x)=\sin^2 x## and it works again wolfram says the integrals are equal to ##3\frac{\pi}{8}##.

Including an additional factor of ##1/2##!

Er yes the integral on the left is ##\frac{3\pi}{8}## and on the right ##\frac{3\pi}{4}##.

So it seems to be true after all, but I fall into an infinite loop if I try to use integration by factors to prove it.

Tried ##f(x)=\cos x## and again it is true, (integrals are now equal to ##\pi## and ##\pi/2##.) so not sure if the requirement for f being positive is of any use.

Proved it by considering ##f## as Fourier series $$f(x)=\sum a_n \sin(nx)+b_n\cos(nx)$$.

Since we are taking the definite integrals over the period interval, many integrals in the resulting sums become zero and only some crucial integrals survive that equal to multiples of pi. The equality to prove goes down to $$\sum (a_n^2+b_n^2)\frac{\pi}{2}$$ in both sides.

Not so easy not so hard either, maybe there is a clever shortcut with integration by factors? Come to think of it I should 've named this thread Fourier series, unless someone can find a clever shortcut using integration by factors.

Delta2 said:

Homework Statement: if ##f(x)## has period ##2\pi## and it is positive and differentiable, prove that $$\int_0^{2\pi} (f(x)\cos x)(f(x)\sin x)'dx=\frac{1}{2}\int_0^{2\pi}f^2(x)dx$$

There's no need for ##f(x)## to be positive:
$$\int_0^{2\pi} (f(x)\cos x)(f(x)\sin x)' \ dx = \int_0^{2\pi} (f(x)\cos x)(f'(x)\sin x + f(x)\cos x) \ dx$$$$= \int_0^{2\pi} f^2(x)\cos^2 x + f(x)f'(x)\cos x\sin x \ dx$$$$= \int_0^{2\pi} f^2(x)\cos^2 x + \frac 1 2 \int_0^{2\pi} (f^2(x))'\cos x\sin x \ dx$$$$= \int_0^{2\pi} f^2(x)\cos^2 x - \frac 1 2 \int_0^{2\pi} (f^2(x))(\cos x\sin x)' \ dx \ \ (\text{by parts})$$$$= \dots$$

Well done @PeroK you found the clever shortcut, there was a solution with integration by parts as my meta-intuition suggested but for some reason I couldn't find it, I thought expanding the derivative is the first step towards hell e hehe.

But kind of fool you too for another reason, you initially thought it was wrong.

Delta2 said:

Homework Statement: if ##f(x)## has period ##2\pi## and it is positive and differentiable, prove that $$\int_0^{2\pi} (f(x)\cos x)(f(x)\sin x)'dx=\frac{1}{2}\int_0^{2\pi}f^2(x)dx$$

Also, there's no requirement for ##f## to be be periodic.

PeroK said:

Also, there's no requirement for ##f## to be be periodic.

There is : you cant set to zero that "thing" (the first term from integration by parts) if f is not periodic but ye just $$f(0)=f(2\pi)$$ is enough if that's what you mean.

Oops yeah you right there is ##\sin(0)=\sin(2\pi)=0## there :D.

I think a huge part of this is to use double angle identities and power reduction formulas. Let me know if I did anything wrong.

##\int_{0}^{2 \pi} \left[ f\left(x\right) \cos x \right] \left[ f\left(x\right) \sin x \right]' \, dx##

##= \int_{0}^{2 \pi} \left[ f\left(x\right) \cos x \right] \left[ f'\left(x\right) \sin x + f\left(x\right) \cos x\right] \, dx##

## = \int_{0}^{2 \pi} f\left(x \right) f' \left(x \right) \cos x \sin x + f^2 \left(x\right) \cos^2 x \,dx##

I'm going to use a double identity ##\sin 2x = 2 \sin x \cos x## and power reduction formula ##\cos^2 x = \frac{1 + \cos 2x}{2}##

## = \int_{0}^{2 \pi} f\left(x \right) f' \left(x \right) \frac{\sin 2x}{2} + f^2 \left(x\right) \frac{1 + \cos 2x}{2} \,dx##

## = \frac{1}{2} \int_{0}^{2 \pi} \left(f'\left(x\right) f\left(x\right) \sin 2x + f^2 \left(x\right) \cos 2x\right) \,dx + \frac{1}{2} \int_{0}^{2 \pi} f^2\left(x\right)\,dx##

Recognize the first integrand (rather the first two) as a derivative of ##\frac{1}{4} f^2 \left( x \right) \sin 2x## by the product rule

##= \frac{1}{4} \int_{0}^{2 \pi} \frac{d}{dx} \left(f^2 \sin 2x \right)\,dx + \frac{1}{2} \int_{0}^{2 \pi} f^2 \left(x \right) \, dx##

Use the fundamental theorem of calculus on the first integral

## = \frac{1}{4} \left. f^2 \left(x\right) \sin 2x \right|_{0}^{2 \pi} + \frac{1}{2} \int_{0}^{2 \pi} f^2 \left(x \right) \, dx##

## = 0 + \frac{1}{2} \int_{0}^{2 \pi} f^2 \left(x \right) \, dx##

##= \frac{1}{2} \int_{0}^{2 \pi} f^2 \left(x \right) \, dx##

@PhDeezNutz looks correct to me well done and you didn't use Integration by Factors, just well known trig identities and the FTC. Thanks!

Delta2 said:

@PhDeezNutz looks correct to me well done and you didn't use Integration by Factors, just well known trig identities and the FTC. Thanks!

Did the problem statement explicitly ask to use integration by factors? If so I might have another go at it later today.

PhDeezNutz said:

Did the problem statement explicitly ask to use integration by factors? If so I might have another go at it later today.

Nope, I just thought that was the most straightforward way to do it.

PhDeezNutz said:

Did the problem statement explicitly ask to use integration by factors? If so I might have another go at it later today.

Note that your method is an alternative way to do integration by parts. One of my grumbles about maths teaching is that integration by parts is presented as some exotic formula. Whereas, I would call it the inverse product rule, which is more self explanatory. From the product rule we have:
$$(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)$$Integrating that equation gives:
$$\int(f(x)g(x))'\ dx= \int f'(x)g(x) \ dx + \int f(x)g'(x) \ dx$$And, applying the FTC gives:
$$f(x)g(x) = \int f'(x)g(x) \ dx + \int f(x)g'(x) \ dx$$And, finally, the parts formula:
$$\int f'(x)g(x) \ dx = f(x)g(x) - \int f(x)g'(x) \ dx$$Note that as long as you remember the product rule, you can forget integration by parts entirely.

PS note that because we have an indefinite integral on both sides of the equation, we do not need an explicit constant of integration.

PeroK said:

Note that as long as you remember the product rule, you can forget integration by parts entirely.

Exactly! IBF is actually product rule +FTC.