Trigonometric Integration / relationship between f(cosx) and f(sinx)

[Zatman]

Homework Statement

Show that:

∫f(sin(x))dx = ∫f(cos(x))dx

where each integral is over the limits [0, $\pi$/2], for a 'well behaved' function f.

2. The attempt at a solution

I have tried relating sin(x) and cos(x) and somehow rearrange one of the integrals to look like the other. Since:

sin(x) = cos(x-$\pi$/2)

I can substitute this into the LHS integral and I am sure this is the correct way to start. It is at this point where I cannot think of any way of moving forward.

Any help would be appreciated.

 

[Millennial]

You are heading the right way. I'll give you two tips. The first is, how are $\displaystyle \int_{0}^{t} f(x)\,dx$ and $\displaystyle \int_{0}^{t} f(t-x)\,dx$ related?

The second tip is, think about the fundamental properties of sine and cosine. There is one particular property that will help you.

 

[Zatman]

Thank you for you reply Millennial.

Using the property that cos(-x) = cos(x) I have got:

int[f(sinx)]dx = int[f(cos(x-$\pi$/2))] = int[f(cos($\pi$/2-x))]

Looking at your first tip one would assume that the relation is that they are equal (since that yields the required result), which is obvious from a sketch of the graphs of cos(x) and cos($\pi$/2-x). But I am not sure how you would derive this more formally?

 

[haruspex]

But I am not sure how you would derive this more formally?

By a change of variable. In the integral of f(t-x).dx, substitute x = t - u wherever x occurs (including the bounds).

 

[Zatman]

Sorry but I am still confused:

I = int(0 to t)[f(t-x)]dx

let u = t - x
then du = -dx
limits become t to 0

I = -int(t to 0)[f(u)]du
I = int(0 to t)[f(u)]du

Now what? If I put x back in I just get what I started with.

 

[haruspex]

I = int(0 to t)[f(u)]du
Now what? If I put x back in I just get what I started with.

The variable of integration is a 'dummy' variable. You can change it to anything you like so long as you do it consistently. If I = int(0 to t)[f(u)]du then it must also be true that I = int(0 to t)[f(x)]dx.

 

[Zatman]

I see, the value of the integral depends only on the limits and the form of the integrand.

Thank you for your help haruspex and Millennial. :)