Gamma function as solution to an integral

[Vey2000]

Homework Statement

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 $\sin x = t$

Homework Equations

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

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.

 

[mighty2000]

After further consideration, I realized that the suggested change of variable was not the most efficient approach to solving this integral. Instead, I tried using the trigonometric identity \sin^2 x = \frac{1-\cos 2x}{2} to rewrite the integral as \int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{1+\frac{1-\cos 2x}{2}}} dx. This simplifies to \int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{\frac{3+\cos 2x}{2}}} dx.

Next, I used the substitution u = \cos 2x, so that du = -2\sin 2x dx and the limits of integration become u(0) = 1 and u(\frac{\pi}{2}) = 0. The integral then becomes \int_1^0 \frac{-2}{\sqrt{\frac{3+u}{2}}} du.

Using the definition of the gamma function, I can express this as \int_1^0 \frac{-2}{\sqrt{\frac{3+u}{2}}} du = -2\int_1^0 \frac{\sqrt{u}}{\sqrt{\frac{3+u}{2}}} du = -2\int_1^0 \frac{u^{\frac{1}{2}-1}}{\left(\frac{3+u}{2}\right)^{\frac{1}{2}}} du = -2\int_1^0 \frac{u^{\frac{1}{2}-1}}{\left(\frac{3+u}{2}\right)^{\frac{1}{2}}} \frac{2}{3+u} \frac{3+u}{2} du.

Simplifying, this becomes -2\int_1^0 \frac{u^{\frac{1}{2}-1}}{\left(\frac{3+u}{2}\right)^{\frac{1}{2}}} \frac{2}{3+u} \frac{3+u}{2} du = -2\int_1^0 \frac{u^{\frac{1}{2}-1}}{\left(\frac{3+u}{2}\right)^{\frac{1}{2}}} \frac{1}{\left(\frac{