`int_(0)^(pi//4)(2sec^(2)x+x^(3)+2)dx`

Answer

To solve the integral \( \int_{0}^{\frac{\pi}{4}} (2 \sec^2 x + x^3 + 2) \, dx \), we can break it down into three separate integrals:

1. \( \int_{0}^{\frac{\pi}{4}} 2 \sec^2 x \, dx \)
2. \( \int_{0}^{\frac{\pi}{4}} x^3 \, dx \)
3. \( \int_{0}^{\frac{\pi}{4}} 2 \, dx \)

Now, let's solve each of these integrals step by step.

### Step 1: Integrate \( 2 \sec^2 x \)

The integral of \( \sec^2 x \) is \( \tan x \). Therefore,

\[ \int 2 \sec^2 x \, dx = 2 \tan x + C \]

Now we evaluate it from \( 0 \) to \( \frac{\pi}{4} \):

\[ \left[ 2 \tan x \right]_{0}^{\frac{\pi}{4}} = 2 \tan\left(\frac{\pi}{4}\right) - 2 \tan(0) = 2 \cdot 1 - 2 \cdot 0 = 2 \]

### Step 2: Integrate \( x^3 \)

The integral of \( x^3 \) is given by:

\[ \int x^3 \, dx = \frac{x^4}{4} + C \]

Now we evaluate it from \( 0 \) to \( \frac{\pi}{4} \):

\[ \left[ \frac{x^4}{4} \right]_{0}^{\frac{\pi}{4}} = \frac{\left(\frac{\pi}{4}\right)^4}{4} - \frac{0^4}{4} = \frac{\frac{\pi^4}{256}}{4} = \frac{\pi^4}{1024} \]

### Step 3: Integrate \( 2 \)

The integral of \( 2 \) is:

\[ \int 2 \, dx = 2x + C \]

Now we evaluate it from \( 0 \) to \( \frac{\pi}{4} \):

\[ \left[ 2x \right]_{0}^{\frac{\pi}{4}} = 2 \cdot \frac{\pi}{4} - 2 \cdot 0 = \frac{\pi}{2} \]

### Step 4: Combine the results

Now, we combine the results of the three integrals:

\[ \int_{0}^{\frac{\pi}{4}} (2 \sec^2 x + x^3 + 2) \, dx = 2 + \frac{\pi^4}{1024} + \frac{\pi}{2} \]

Thus, the final answer is:

\[ \int_{0}^{\frac{\pi}{4}} (2 \sec^2 x + x^3 + 2) \, dx = 2 + \frac{\pi^4}{1024} + \frac{\pi}{2} \]