Integral of f(x): Get Answer Here

[Hamal_Arietis]

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

 

[pasmith]

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 <br /> 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?

 

[Hamal_Arietis]

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:
$$\int_0^{\frac{\pi}{a}}f(t)cos^2(at)dt=\int_0^{\frac{\pi}{a}}f(t)sin^2(at)dt \Rightarrow f(t)cos^2(at)=f(t)sin^2(at)\Leftrightarrow f(t)=0 ∨ at=\frac{\pi}{4}+k\pi (k∈Z)$$
So the answer is:
$$f(x)=\frac{\pi}{a}+1$$
Is this right?

 

[Ray Vickson]

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]

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]

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:
$$\int_0^{\frac{\pi}{a}}f(t)cos^2(at)dt=\int_0^{\frac{\pi}{a}}f(t)sin^2(at)dt \Rightarrow f(t)cos^2(at)=f(t)sin^2(at)\Leftrightarrow f(t)=0 ∨ at=\frac{\pi}{4}+k\pi (k∈Z)$$
So the answer is:
$$f(x)=\frac{\pi}{a}+1$$
Is this right?

No, it is not right. I suggest you go back and look at post #5.

 

[Hamal_Arietis]

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:
$$\int_0^{\frac{\pi}{a}}f(t)cos^2(at)dt=\int_0^{\frac{\pi}{a}}f(t)sin^2(at)dt$$

 

[Hamal_Arietis]

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]

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]

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.

 

[Hamal_Arietis]

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]

$$A=\int_0^{\frac{\pi}{a}}f(t)\cos at\,dt$$
$$A=\int_0^{\frac{\pi}{a}}( 1 + A \cos 2at + B \sin 2at)cos(at)dt$$
$$A=\frac{4B}{3a}$$
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.