`int_(0)^(pi//4) cos^(2) x dx`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{4}} \cos^2 x \, dx \), we can use the trigonometric identity for \( \cos^2 x \):

\[ \cos^2 x = \frac{1 + \cos(2x)}{2} \]

Now, we can rewrite the integral using this identity:

\[ I = \int_{0}^{\frac{\pi}{4}} \cos^2 x \, dx = \int_{0}^{\frac{\pi}{4}} \frac{1 + \cos(2x)}{2} \, dx \]

Next, we can separate the integral:

\[ I = \frac{1}{2} \int_{0}^{\frac{\pi}{4}} 1 \, dx + \frac{1}{2} \int_{0}^{\frac{\pi}{4}} \cos(2x) \, dx \]

Now, we can evaluate each integral separately.

1. The first integral is straightforward:

\[ \int_{0}^{\frac{\pi}{4}} 1 \, dx = \left[ x \right]_{0}^{\frac{\pi}{4}} = \frac{\pi}{4} - 0 = \frac{\pi}{4} \]

2. The second integral requires a simple substitution. The integral of \( \cos(2x) \) is:

\[ \int \cos(2x) \, dx = \frac{1}{2} \sin(2x) + C \]

Thus, we evaluate:

\[ \int_{0}^{\frac{\pi}{4}} \cos(2x) \, dx = \left[ \frac{1}{2} \sin(2x) \right]_{0}^{\frac{\pi}{4}} = \frac{1}{2} \sin\left(2 \cdot \frac{\pi}{4}\right) - \frac{1}{2} \sin(0) = \frac{1}{2} \sin\left(\frac{\pi}{2}\right) - 0 = \frac{1}{2} \cdot 1 = \frac{1}{2} \]

Now we can substitute back into our expression for \( I \):

\[ I = \frac{1}{2} \left( \frac{\pi}{4} \right) + \frac{1}{2} \left( \frac{1}{2} \right) = \frac{\pi}{8} + \frac{1}{4} \]

To combine these fractions, we need a common denominator:

\[ \frac{1}{4} = \frac{2}{8} \]

So,

\[ I = \frac{\pi}{8} + \frac{2}{8} = \frac{\pi + 2}{8} \]

Thus, the final answer is:

\[ \int_{0}^{\frac{\pi}{4}} \cos^2 x \, dx = \frac{\pi + 2}{8} \]