Compute the integral of $ cos^2(t) $ on the unit circle

To compute the integral of $\cos^2(t)$ on the unit circle, we can use the double-angle identity for cosine:

$\cos^2(t) = \frac{1 + \cos(2t)}{2}$

Now, integrate:

$\int_0^{2\pi} \cos^2(t) \, dt = \int_0^{2\pi} \frac{1 + \cos(2t)}{2} \, dt$

Separate the integral:

$= \frac{1}{2} \int_0^{2\pi} (1 + \cos(2t)) \, dt$

Split it into two integrals:

$= \frac{1}{2} \left( \int_0^{2\pi} 1 \, dt + \int_0^{2\pi} \cos(2t) \, dt \right)$

The first integral is straightforward:

$= \frac{1}{2} \left( 2\pi + 0 \right)$

The second integral of cosine over a full period is zero:

$= \pi$

To compute the integral of $cos^2(t)$ on the unit circle, use the identity:

$cos^2(t) = frac{1 + cos(2t)}{2}$

Then:

$int_0^{2pi} cos^2(t) , dt = int_0^{2pi} frac{1 + cos(2t)}{2} , dt$

Separate it:

$= frac{1}{2} left( int_0^{2pi} 1 , dt + int_0^{2pi} cos(2t) , dt
ight)$

We get:

$= frac{1}{2} left( 2pi + 0
ight)$

The result:

$= pi$

To find the integral of $cos^2(t)$ on the unit circle, use:

$cos^2(t) = frac{1 + cos(2t)}{2}$

Integrate:

$int_0^{2pi} cos^2(t) , dt = frac{1}{2} left( 2pi + 0
ight) = pi$

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