Integral and Minimum value of a function

[grapz]

A function f is defined by:

f ( x ) = [int] from 0 -> pi cos t * cos ( x - t) dt 0 <_ x <_ 2pi

Find the minimum value of f

I first use the trig identity cosAcosB = 1/2 [ cos( A+B) +cos(A-B)]

So the integrand simplies into this:

1/2 [int] 0--> pi [ cos x + cos(2t - x)] dt

I am stuck on how to solve for this integral because of the two variables.

 

[galactus]

If I am reading it correctly, you have \(\displaystyle \int_{0}^{\pi}[cos(t)cos(x-t)]dt\)

Treat x as a constant. Then integrate wrt t

Integrate and find the minimum value over the interval \(\displaystyle [0,2\pi]\)

Cosine ranges from -1 to 1. \(\displaystyle cos(\pi)=-1\)

Using what you have, what would be the minimum value?.

You can always graph the function.

 

[Dr. Flim-Flam]

f(x) = integral [cos(t)cos(x-t),t,0,Pi] = [(Pi/2)cos(x)}

f"(x) = -(Pi/2)sin(x) = 0. x = 0, Pi, 2(Pi).

f(0) = Pi/2, f(Pi) = -Pi/2, f[2(Pi)] = Pi/2, hence min = (Pi, -Pi/2)

Note; Use identity cos(x-t) = cos(x)cos(t) + sin(x)sin(t) and then split the integral (Just one of many ways).