Show that integral from 0 to π/2 of √cos(θ) dθ =

Steps to Solve

1. Set up the integral

We need to evaluate the integral

$$ I = \int_0^{\frac{\pi}{2}} \sqrt{\cos(x)} , dx. $$

2. Use substitution

We will use the substitution ( u = \cos(x) ). Then, the derivative ( du = -\sin(x) , dx ) or ( dx = -\frac{du}{\sin(x)} ). Since ( \sin^2(x) + \cos^2(x) = 1 ), we have:

$$ \sin(x) = \sqrt{1 - u}. $$

Now, we need to change the limits of integration.

• When ( x = 0 ), ( u = \cos(0) = 1 ).
• When ( x = \frac{\pi}{2} ), ( u = \cos\left(\frac{\pi}{2}\right) = 0 ).

Therefore, the integral becomes:

$$ I = \int_1^0 \sqrt{u} \left(-\frac{du}{\sqrt{1-u}}\right) = \int_0^1 \frac{\sqrt{u}}{\sqrt{1-u}} , du. $$

3. Recognize the form of the integral

The integral

$$ \int_0^1 \frac{\sqrt{u}}{\sqrt{1-u}} , du $$

can be solved using the Beta function or recognizing it as a known integral:

$$ I = \int_0^1 u^{\frac{1}{2}} (1-u)^{-\frac{1}{2}} , du $$

which corresponds to

$$ B\left( \frac{3}{2}, \frac{1}{2} \right), $$

where the Beta function is defined as

$$ B(x, y) = \int_0^1 t^{x-1} (1-t)^{y-1} , dt. $$

4. Use the properties of the Beta function

Using the property ( B(x, y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)} ):

For our integral, we have:

$$ B\left( \frac{3}{2}, \frac{1}{2} \right) = \frac{\Gamma\left(\frac{3}{2}\right) \Gamma\left(\frac{1}{2}\right)}{\Gamma\left(2\right)}. $$

Knowing that ( \Gamma\left(\frac{1}{2}\right) = \sqrt{\pi} ) and ( \Gamma\left(2\right) = 1 ):

5. Calculate the value

Since ( \Gamma\left(\frac{3}{2}\right) = \frac{1}{2} \sqrt{\pi} ):

Now we can find:

$$ B\left( \frac{3}{2}, \frac{1}{2} \right) = \frac{\frac{1}{2} \sqrt{\pi} \cdot \sqrt{\pi}}{1} = \frac{1}{2} \pi. $$

Thus, we find:

$$ I = \frac{1}{2} \pi. $$