# Gamma function as solution to an integral

1. Feb 26, 2007

### Vey2000

1. The problem statement, all variables and given/known data

Calculate $$\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{1+\sin^2 x}} dx$$ expressing its solution in terms of the gamma function. It's suggested to first use the change of variable $\sin x = t$

2. Relevant equations

The gamma function is defined as $$p>0, \Gamma(p)=\int_0^\infty x^{p-1} e^{-x} dx$$

3. The attempt at a solution

After using the suggested change of variable the integral is transformed into $$\int_0^1 \frac{dt}{\sqrt{1-t^4}}$$

I then considered using $1-t^4=e^{2y}$ such that the integrand, without yet considering the factor from transforming the differential, would become $e^{-y}$, with the hope that the differential would provide the necessary factor to fit the definition of the gamma function. Unfortunately, the differential turns out to be $$dt=\frac{-e^{2y}dy}{2{\left(e^{2y}-1\right)}^\frac{3}{4}}$$ which obviously isn't simplifying things at all.

Several more variations along this line of thinking followed with similar results.

Also tried $1-t^4=y^a e^{b y}$ which results in $$\int \frac{-(a y^{a/2-1}+b y^{a/2})e^{by/2}dy}{4(1-y^ae^{by})^\frac{3}{4}}$$ which still isn't getting close to the form of the gamma function.

Last edited: Feb 26, 2007