Find `int_(0)^(pi//2)(1-cos theta)^(1//2)d theta`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} (1 - \cos \theta)^{\frac{1}{2}} d\theta \), we can follow these steps:

### Step 1: Simplify the integrand
We know that \( 1 - \cos \theta = 2 \sin^2\left(\frac{\theta}{2}\right) \) from the half-angle identity. Thus, we can rewrite the integrand:

\[ (1 - \cos \theta)^{\frac{1}{2}} = \left(2 \sin^2\left(\frac{\theta}{2}\right)\right)^{\frac{1}{2}} = \sqrt{2} \sin\left(\frac{\theta}{2}\right) \]

### Step 2: Change the variable
Let’s make a substitution to simplify the integral. Set \( u = \frac{\theta}{2} \), which means \( \theta = 2u \) and \( d\theta = 2 du \). The limits change accordingly: when \( \theta = 0 \), \( u = 0 \) and when \( \theta = \frac{\pi}{2} \), \( u = \frac{\pi}{4} \).

Now, substituting into the integral gives:

\[ I = \int_{0}^{\frac{\pi}{4}} \sqrt{2} \sin(u) \cdot 2 \, du = 2\sqrt{2} \int_{0}^{\frac{\pi}{4}} \sin(u) \, du \]

### Step 3: Evaluate the integral
The integral of \( \sin(u) \) is \( -\cos(u) \). Thus, we can evaluate:

\[ \int \sin(u) \, du = -\cos(u) \]

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

\[ \int_{0}^{\frac{\pi}{4}} \sin(u) \, du = -\cos\left(\frac{\pi}{4}\right) - (-\cos(0)) = -\frac{1}{\sqrt{2}} + 1 = 1 - \frac{1}{\sqrt{2}} \]

### Step 4: Substitute back into the integral
Now substitute back into the expression for \( I \):

\[ I = 2\sqrt{2} \left(1 - \frac{1}{\sqrt{2}}\right) = 2\sqrt{2} - 2 \]

### Final Result
Thus, the value of the integral is:

\[ I = 2\sqrt{2} - 2 \]