# Find the integral

## Homework Statement

Find f(x) if:(a∈R)
$$f(x)=\int_0^{\frac{\pi}{a}}f(t)cos(at-2ax)dt+1$$

## Homework Equations

$$f(x)=\int_{a}^{b}f(t)dt=F(a)-F(b)$$
$$\int{udv}=uv-\int{vdu}$$

## The Attempt at a Solution

I tried to use the fundamental of calculus and integration by parts but they don't have answer. Because cos(at-2ax) confuses me.
Thank for helping

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pasmith
Homework Helper

## Homework Statement

Find f(x) if:(a∈R)
$$f(x)=\int_0^{\frac{\pi}{a}}f(t)cos(at-2ax)dt+1$$

## Homework Equations

$$f(x)=\int_{a}^{b}f(t)dt=F(a)-F(b)$$
$$\int{udv}=uv-\int{vdu}$$

## The Attempt at a Solution

I tried to use the fundamental of calculus and integration by parts but they don't have answer. Because cos(at-2ax) confuses me.
Thank for helping
The fundamental theorem may not help, since you don't know an antiderivative of $f$.

Instead use $\cos(at - 2ax) = \cos(at)\cos(2ax) + \sin(at)\sin(2ax)$ to obtain $$f(x) = 1 + \cos(2ax)\int_0^{\pi/a}f(t)\cos(at)\,dt + \sin(2ax)\int_0^{\pi/a}f(t)\sin(at)\,dt.$$ Now set $A = \int_0^{\pi/a} f(t) \cos(at)\,dt$, $B = \int_0^{\pi/a} f(t) \sin(at)\,dt$ so that $f(t) = 1 + A\cos(2at) + B\sin(2at)$ and you can obtain a pair of simultaneous equations to be solved for $A$ and $B$.

• Hamal_Arietis
so that ##f(t)=1+Acos(2at)+Bsin(2at)f(t)=1+Acos⁡(2at)+Bsin⁡(2at)f(t) = 1 + A\cos(2at) + B\sin(2at) ##and you can obtain a pair of simultaneous equations to be solved for A and B.
Thanks, I understood. Another equation is:
$$f(t)=\int_0^{\frac{\pi}{a}}f(t)cos(at)dt+1=A+1$$
But we must find f(t), A, B while we have 2 equations?

Oh, I see:
$$A=Acos(2at)+Bsin(2at)$$
$$A(1-cos(2at))=Bsin(2at)$$
$$2Acos^2(at)=2Bsin(at)cos(at)$$
$$cos(at)(Acos(at)-Bsin(at))=0$$
So we have:
$$\int_0^{\frac{\pi}{a}}f(t)cos^2(at)dt=\int_0^{\frac{\pi}{a}}f(t)sin^2(at)dt \Rightarrow f(t)cos^2(at)=f(t)sin^2(at)\Leftrightarrow f(t)=0 ∨ at=\frac{\pi}{4}+k\pi (k∈Z)$$
$$f(x)=\frac{\pi}{a}+1$$
Is this right?

Last edited:
Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Find f(x) if:(a∈R)
$$f(x)=\int_0^{\frac{\pi}{a}}f(t)cos(at-2ax)dt+1$$

## Homework Equations

$$f(x)=\int_{a}^{b}f(t)dt=F(a)-F(b)$$
$$\int{udv}=uv-\int{vdu}$$

## The Attempt at a Solution

I tried to use the fundamental of calculus and integration by parts but they don't have answer. Because cos(at-2ax) confuses me.
Thank for helping
Try looking at ##f'(x)## and ##f''(x)##, to see if ##f(x)## obeys a simple differential equation.

vela
Staff Emeritus
Homework Helper
Oh, I see:
$$A=Acos(2at)+Bsin(2at)$$
Where did this come from?

$$\int_0^{\frac{\pi}{a}}f(t)cos^2(at)dt=\int_0^{\frac{\pi}{a}}f(t)sin^2(at)dt \Rightarrow f(t)cos^2(at)=f(t)sin^2(at)$$
No. You can't infer that because the two integrals are equal the two integrands are equal.

What pasmith suggested was you have an expression for f(t) in terms of A and B. Plug that into the expressions that define what A and B are and evaluate the integrals. You'll end up with two equations which you can solve for A and B.

Ray Vickson
Homework Helper
Dearly Missed
Oh, I see:
$$A=Acos(2at)+Bsin(2at)$$
$$A(1-cos(2at))=Bsin(2at)$$
$$2Acos^2(at)=2Bsin(at)cos(at)$$
$$cos(at)(Acos(at)-Bsin(at))=0$$
So we have:
$$\int_0^{\frac{\pi}{a}}f(t)cos^2(at)dt=\int_0^{\frac{\pi}{a}}f(t)sin^2(at)dt \Rightarrow f(t)cos^2(at)=f(t)sin^2(at)\Leftrightarrow f(t)=0 ∨ at=\frac{\pi}{4}+k\pi (k∈Z)$$
$$f(x)=\frac{\pi}{a}+1$$
Is this right?
No, it is not right. I suggest you go back and look at post #5.

Where did this come from?

No. You can't infer that because the two integrals are equal the two integrands are equal.

What pasmith suggested was you have an expression for f(t) in terms of A and B. Plug that into the expressions that define what A and B are and evaluate the integrals. You'll end up with two equations which you can solve for A and B.
Both equations are:
$$f(t)=A+1$$
$$f(t)=1+Acos(2at)+Bsin(2at)$$
So ##A=Acos(2at)+Bsin(2at)##
and I transfer it into:
$$\int_0^{\frac{\pi}{a}}f(t)cos^2(at)dt=\int_0^{\frac{\pi}{a}}f(t)sin^2(at)dt$$

Last edited:
Try looking at ##f'(x)## and ##f''(x)##, to see if ##f(x)## obeys a simple differential equation.
I try to find f'(x) by the fundamental of calculus before but it don't have answer

Ray Vickson
Homework Helper
Dearly Missed
I try to find f'(x) by the fundamental of calculus before but it don't have answer
You can find a formula for
$$\frac{d}{dx} \int_a^b h(x,t) \, dt$$
It has nothing to do with the fundamental theorem of calculus.

• Hamal_Arietis
You can find a formula for
$$\frac{d}{dx} \int_a^b h(x,t) \, dt$$
It has nothing to do with the fundamental theorem of calculus.
Sorry, I don't understand your idea. Can you give me some document about it?. This problem is an examination for high school students to entrance university. I have learned that:
If ##F(x)=\int_a^xf(t)dt## then:
$$\frac{d}{dx}F(x)=\frac{d}{dx}\int_a^xf(t)dt=f(x)$$

vela
Staff Emeritus
Homework Helper
Both equations are:
$$f(t)=A+1$$
Where did this come from? Note the lefthand side depends on ##t##, but the righthand side is a constant. It doesn't look right to me.

You have ##f(t) = 1 + A \cos 2at + B \sin 2at## where ##A = \int_0^{\frac{\pi}{a}}f(t)\cos at\,dt##. Plug the expression for ##f(t)## into the definition of ##A## and evaluate the integral.

Where did this come from? Note the lefthand side depends on ##t##, but the righthand side is a constant. It doesn't look right to me.

You have ##f(t) = 1 + A \cos 2at + B \sin 2at## where ##A = \int_0^{\frac{\pi}{a}}f(t)\cos at\,dt##. Plug the expression for ##f(t)## into the definition of ##A## and evaluate the integral.
$$A=\int_0^{\frac{\pi}{a}}f(t)\cos at\,dt$$
$$A=\int_0^{\frac{\pi}{a}}( 1 + A \cos 2at + B \sin 2at)cos(at)dt$$
$$A=\frac{4B}{3a}$$
And $$B=\int_0^{\frac{\pi}{a}}( 1 + A \cos 2at + B \sin 2at)sin(at)dt$$
$$B=\frac{2}{a}+\frac{2A}{3a}=\frac{6+2A}{3a}$$
Then solve both equation:
$$A=\frac{24}{9a^2-8};B=\frac{18a}{9a^2-8}$$

Ray Vickson
Homework Helper
Dearly Missed
$$A=\int_0^{\frac{\pi}{a}}f(t)\cos at\,dt$$
$$A=\int_0^{\frac{\pi}{a}}( 1 + A \cos 2at + B \sin 2at)cos(at)dt$$
$$A=\frac{4B}{3a}$$
And $$B=\int_0^{\frac{\pi}{a}}( 1 + A \cos 2at + B \sin 2at)sin(at)dt$$
$$B=\frac{2}{a}+\frac{2A}{3a}=\frac{6+2A}{3a}$$
Then solve both equation:
$$A=\frac{24}{9a^2-8};B=\frac{18a}{9a^2-8}$$
Almost OK: the ##B##-equation should be
$$B=\frac{2}{a} - \frac{2A}{3a},$$
which changes the final solution a bit.

• Hamal_Arietis