`int_(0)^(pi//2) (a cos^(2) x+b sin^(2) x) dx`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} (a \cos^2 x + b \sin^2 x) \, dx \), we will follow these steps:

### Step 1: Rewrite the integrand using trigonometric identities
We know the following identities:
\[ \cos^2 x = \frac{1 + \cos 2x}{2} \quad \text{and} \quad \sin^2 x = \frac{1 - \cos 2x}{2} \]
Substituting these into the integral, we get:
\[ I = \int_{0}^{\frac{\pi}{2}} \left( a \left(\frac{1 + \cos 2x}{2}\right) + b \left(\frac{1 - \cos 2x}{2}\right) \right) \, dx \]

### Step 2: Simplify the integrand
Combining the terms inside the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} \left( \frac{a + b}{2} + \frac{a - b}{2} \cos 2x \right) \, dx \]

### Step 3: Split the integral
We can split the integral into two parts:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{a + b}{2} \, dx + \int_{0}^{\frac{\pi}{2}} \frac{a - b}{2} \cos 2x \, dx \]

### Step 4: Evaluate the first integral
The first integral is straightforward:
\[ \int_{0}^{\frac{\pi}{2}} \frac{a + b}{2} \, dx = \frac{a + b}{2} \cdot \left[ x \right]_{0}^{\frac{\pi}{2}} = \frac{a + b}{2} \cdot \frac{\pi}{2} = \frac{(a + b)\pi}{4} \]

### Step 5: Evaluate the second integral
For the second integral, we have:
\[ \int_{0}^{\frac{\pi}{2}} \frac{a - b}{2} \cos 2x \, dx \]
Using the integral of cosine:
\[ \int \cos kx \, dx = \frac{1}{k} \sin kx \]
Thus,
\[ \int_{0}^{\frac{\pi}{2}} \cos 2x \, dx = \left[ \frac{1}{2} \sin 2x \right]_{0}^{\frac{\pi}{2}} = \frac{1}{2} (\sin \pi - \sin 0) = \frac{1}{2} (0 - 0) = 0 \]

### Step 6: Combine the results
Putting it all together, we find:
\[ I = \frac{(a + b)\pi}{4} + 0 = \frac{(a + b)\pi}{4} \]

### Final Result
Thus, the value of the integral is:
\[ \int_{0}^{\frac{\pi}{2}} (a \cos^2 x + b \sin^2 x) \, dx = \frac{(a + b)\pi}{4} \]