Question

$\int_{0}^{\pi/2} cos^{1/2} x \ dx$

Answer 1

$\frac{2 \sqrt{\pi} \Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}$

Answer 2

To evaluate the definite integral $\int_{0}^{\pi/2} \cos^{1/2} x \ dx$, we can use the properties of Beta and Gamma functions.

The integral is of the form $\int_{0}^{\pi/2} \sin^m x \cos^n x \ dx$.
In our case, $m=0$ and $n=1/2$.

The general formula for this type of integral in terms of Gamma functions is: $\int_{0}^{\pi/2} \sin^m x \cos^n x \ dx = \frac{\Gamma\left(\frac{m+1}{2}\right) \Gamma\left(\frac{n+1}{2}\right)}{2 \Gamma\left(\frac{m+n+2}{2}\right)}$

Substitute $m=0$ and $n=1/2$ into the formula: $\int_{0}^{\pi/2} \cos^{1/2} x \ dx = \frac{\Gamma\left(\frac{0+1}{2}\right) \Gamma\left(\frac{1/2+1}{2}\right)}{2 \Gamma\left(\frac{0+1/2+2}{2}\right)}$ $= \frac{\Gamma\left(\frac{1}{2}\right) \Gamma\left(\frac{3/2}{2}\right)}{2 \Gamma\left(\frac{5/2}{2}\right)}$ $= \frac{\Gamma\left(\frac{1}{2}\right) \Gamma\left(\frac{3}{4}\right)}{2 \Gamma\left(\frac{5}{4}\right)}$

Now, we use the following properties of the Gamma function:

1. $\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$
2. $\Gamma(z+1) = z \Gamma(z)$

Applying the second property to $\Gamma\left(\frac{5}{4}\right)$: $\Gamma\left(\frac{5}{4}\right) = \Gamma\left(1 + \frac{1}{4}\right) = \frac{1}{4} \Gamma\left(\frac{1}{4}\right)$

Substitute these values back into the expression: $\int_{0}^{\pi/2} \cos^{1/2} x \ dx = \frac{\sqrt{\pi} \Gamma\left(\frac{3}{4}\right)}{2 \left(\frac{1}{4} \Gamma\left(\frac{1}{4}\right)\right)}$ $= \frac{\sqrt{\pi} \Gamma\left(\frac{3}{4}\right)}{\frac{1}{2} \Gamma\left(\frac{1}{4}\right)}$ $= \frac{2 \sqrt{\pi} \Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}$

This is the most simplified form of the integral in terms of Gamma functions. The values of $\Gamma(1/4)$ and $\Gamma(3/4)$ are not elementary and are usually left in this form in competitive exams unless specific numerical approximations or further properties are required.