To evaluate the integral

$\int_0^{\frac{\pi}{2}} \cos^2 x \, dx,$

we can use the trigonometric identity for $\cos^2 x$:

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

Thus, we can rewrite the integral as:

$\int_0^{\frac{\pi}{2}} \cos^2 x \, dx = \int_0^{\frac{\pi}{2}} \frac{1 + \cos(2x)}{2} \, dx.$

This simplifies to:

$\int_0^{\frac{\pi}{2}} \cos^2 x \, dx = \frac{1}{2} \int_0^{\frac{\pi}{2}} (1 + \cos(2x)) \, dx.$

Now we can split the integral:

$= \frac{1}{2} \left( \int_0^{\frac{\pi}{2}} 1 \, dx + \int_0^{\frac{\pi}{2}} \cos(2x) \, dx \right).$

1. First Integral:

   $\int_0^{\frac{\pi}{2}} 1 \, dx = \left[ x \right]_0^{\frac{\pi}{2}} = \frac{\pi}{2}.$

2. Second Integral:

   $\int_0^{\frac{\pi}{2}} \cos(2x) \, dx.$

   To solve this, let’s find the antiderivative of $\cos(2x)$:

   $\int \cos(2x) \, dx = \frac{\sin(2x)}{2}.$

   Evaluating this from $0$ to $\frac{\pi}{2}$:

   $\int_0^{\frac{\pi}{2}} \cos(2x) \, dx = \left[ \frac{\sin(2x)}{2} \right]_0^{\frac{\pi}{2}} = \frac{\sin(\pi)}{2} - \frac{\sin(0)}{2} = 0.$

Therefore, the integral simplifies to:

$\int_0^{\frac{\pi}{2}} \cos^2 x \, dx = \frac{1}{2} \left( \frac{\pi}{2} + 0 \right) = \frac{\pi}{4}.$

Final Answer:

$\int_0^{\frac{\pi}{2}} \cos^2 x \, dx = \frac{\pi}{4}.$

Would you like further details or have any questions?

Here are some related questions to explore:

1. How can we solve integrals involving $\sin^2 x$ in a similar way?
2. What is the integral of $\cos^2 x$ over a different interval, like from $0$ to $\pi$?
3. Can we use a different identity to evaluate (\int \cos^2