$\int_{0}^{\pi /2} (\cos x)^{2.5} \mathrm dx$ $$\int_{0}^{\pi /2} (\cos x)^{2.5} \mathrm dx$$
I tried changing to sine and substituting, but it's getting complex. Help.
 A: It is a difficult problem because you face elliptic integrals.
$$\int\cos ^{\frac{5}{2}}(x)\,dx=\frac{2}{5} \left(3 E\left(\left.\frac{x}{2}\right|2\right)+\sin (x) \cos^{\frac{3}{2}}(x)\right)$$ making
$$\int_0^{\frac \pi 2}\cos ^{\frac{5}{2}}(x)\,dx=\frac{3}{5} \sqrt{2} \left(2
   E\left(\frac{1}{2}\right)-K\left(\frac{1}{2}\right)\right)=\frac{\sqrt{\pi } \,\Gamma \left(\frac{7}{4}\right)}{2\, \Gamma
   \left(\frac{9}{4}\right)}$$
What you could do is to use the Taylor series expansion of $\cos(x)$ and use the binomial theorem to get
$$\cos ^{\frac{5}{2}}(x)=1-\frac{5 x^2}{4}+\frac{55 x^4}{96}-\frac{139 x^6}{1152}+\frac{1709
   x^8}{129024}-\frac{22201 x^{10}}{23224320}+\frac{233771
   x^{12}}{6131220480}+O\left(x^{13}\right)$$ and integtermwise. Using this expansion and the given bounds, you should arrive to $\approx 0.719173$ while the exact value should be $\approx 0.718884$.
In fact, integrals of the type
$$\int\cos ^{a}(x)\,dx=-\frac{\sin (x) \cos ^{a+1}(x)}{(a+1)
   \sqrt{\sin ^2(x)}} \,
   _2F_1\left(\frac{1}{2},\frac{a+1}{2};\frac{a+3}{2};\cos ^2(x)\right)$$ where appears the gaussian hypergeometric function which simplifies only if $a$ is integer.
$$\int_0^{\frac \pi 2}\cos ^{a}(x)\,dx=\frac{\sqrt{\pi }\,\Gamma \left(\frac{a+1}{2}\right)}{2\, \Gamma\left(\frac{a+2}{2}\right)}$$
A: Using the beta function gives
$$\int_0^{\pi/2}(\cos x)^{5/2}\,dx=\frac{B(1/2,7/4)}2=\frac{\Gamma(1/2)\Gamma(7/4)}{2\Gamma(9/4)}=\frac{\sqrt\pi\,\Gamma(7/4)}{2\Gamma(9/4)}.$$
