Evaluate the following :
`int_(-pi//2)^(pi//2) cos theta d theta`

Answer

To evaluate the integral

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos \theta \, d\theta, \]

we can follow these steps:

### Step 1: Identify the properties of the function
The function \(\cos \theta\) is an even function. This means that \(\cos(-\theta) = \cos(\theta)\).

### Step 2: Use the property of even functions
For an even function, the integral from \(-a\) to \(a\) can be simplified as follows:

\[ \int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx. \]

In our case, we can rewrite the integral as:

\[ I = 2 \int_{0}^{\frac{\pi}{2}} \cos \theta \, d\theta. \]

### Step 3: Evaluate the integral from 0 to \(\frac{\pi}{2}\)
Now we need to compute the integral:

\[ \int_{0}^{\frac{\pi}{2}} \cos \theta \, d\theta. \]

The integral of \(\cos \theta\) is \(\sin \theta\). Therefore, we have:

\[ \int \cos \theta \, d\theta = \sin \theta. \]

### Step 4: Apply the limits
Now we apply the limits from \(0\) to \(\frac{\pi}{2}\):

\[ \int_{0}^{\frac{\pi}{2}} \cos \theta \, d\theta = \sin \left( \frac{\pi}{2} \right) - \sin(0). \]

Calculating this gives:

\[ \sin \left( \frac{\pi}{2} \right) = 1 \quad \text{and} \quad \sin(0) = 0. \]

Thus,

\[ \int_{0}^{\frac{\pi}{2}} \cos \theta \, d\theta = 1 - 0 = 1. \]

### Step 5: Multiply by 2
Now, substituting back into our equation for \(I\):

\[ I = 2 \cdot 1 = 2. \]

### Final Result
The value of the integral is:

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos \theta \, d\theta = 2. \]

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