Reducing an explicit integral to beta function values

$\int_{0}^{2} \frac{d x}{\sqrt[3]{x^{2}(2-x)}}$;

$\int_{-1}^{2} \frac{d x}{\sqrt[4]{(2-x)(1+x)^{3}}}$;

$\int_{-2}^{2} \frac{d x}{\sqrt[4]{(2+x)^{3}(2-x)}}$;

$\int_{0}^{2} \frac{d x}{\sqrt[5]{x^{3}(2-x)^{2}}}$;

$\int_{1}^{2} \sqrt[3]{(2-x)^{2}(x-1)} d x$;

$\int_{0}^{3} \frac{d x}{\sqrt[5]{x^{3}(3-x)^{3}}}$.

$\int_{1}^{2} \sqrt{\frac{x-1}{2-x}} \frac{d x}{(x+3)^{2}}$;

$\int_{0}^{1} \sqrt{\frac{1-x}{x}} \frac{d x}{(x+2)^{2}}$;

$\int_{0}^{1} \frac{d x}{(x+1) \sqrt[3]{x^{2}-x^{3}}}$;

$\int_{0}^{1} \frac{x d x}{(2-x) \sqrt[3]{x^{2}(1-x)}}$;

$\int_{0}^{1} \cdot \sqrt{x(1-x)} \cdot d x$;

$\int_{0}^{1} \frac{\sqrt[4]{x(1-x)^{3}}}{(x+1)^{3}} d x$

$\int_{0}^{+\infty } \frac{d x}{1+x^{3}}$;

$\int_{0}^{+\infty } \frac{\sqrt[4]{x}}{(1+x)^{2}} d x$;

$\int_{0}^{a} x^{2} \sqrt{a^{2}-x^{2}} d x$;

$\int_{-\infty }^{+\infty } \frac{x e^{x / 2}}{e^{2 x}+1} d x$;

$\int_{0}^{+\infty } \frac{x^{\alpha -1}}{1+x^{\beta }} d x, 0 < \alpha < \beta$

$\int_{0}^{1} \frac{x^{1-\alpha }(1-x)^{\alpha }}{(x+1)^{3}} d x,-1 < \alpha < 2$.

$\int_{0}^{+\infty } \frac{x^{\alpha -1}}{(1+x)^{2}} d x, 0 < \alpha < 2$;

$\int_{0}^{+\infty } \frac{x^{p+1}}{\left(1+x^{2}\right)^{2}} d x,-2 < p < 2$;

$\int_{0}^{\pi / 2} \frac{\operatorname {tg}^{\alpha } x}{(\sin x+\cos x)^{2}} d x, 0 < \alpha < 1$;

$\int_{0}^{1} x \sqrt[3]{1-x^{3}} d x$;

$\int_{0}^{1} \frac{x^{2 n}}{\sqrt[3]{x\left(1-x^{2}\right)}} d x, n \in N$;

$\int_{0}^{+\infty } \frac{x d x}{\left(1+x^{3}\right)^{2}}$;

$\int_{0}^{+\infty } \frac{x^{2 \alpha -1}}{1+x^{2}} d x, 0 < \alpha < 1$;

$\int_{0}^{\pi / 2} \sin^{4} x \cos^{6} x d x$;

$\int_{0}^{\pi / 2} \operatorname {tg}^{2 \alpha -1} x d x, 0 < \alpha < 1$.

$\int_{0}^{1} \frac{d x}{\sqrt[n]{1-x^{\alpha }}}, \alpha > 0$;

$\int_{0}^{1} x^{\alpha }\left(1-x^{\beta }\right)^{\gamma } d x, \alpha > -1, \quad \beta > 0, \quad \gamma > -1$;

$\int_{0}^{a} x^{\alpha -1}(a-x)^{\beta -1} d x, \quad a > 0, \quad \alpha > 0, \quad \beta > 0$;

$\int_{0}^{+\infty } \frac{x^{\alpha -1}}{(1+x)^{\beta }} d x, 0 < \alpha < \beta ;$

$\int_{0}^{1} \frac{x^{3 \alpha }}{\sqrt[3]{1-x^{3}}} d x, \quad \alpha > -\frac{1}{3}$;

$\int_{0}^{+\infty } \frac{x^{\alpha }}{\left(a+b x^{\beta }\right)^{p}} d x, a > 0, \quad b > 0, \quad \beta > 0, \quad 0 < \frac{\alpha +1}{\beta } < p$.

$\int_{0}^{\pi } \frac{\sin^{p} x}{1+\cos x} d x, \quad p > 1$;

$\int_{0}^{\pi / 2} \frac{\sin^{\alpha -1} x \cos^{\beta -1} x}{(\sin x+\cos x)^{\alpha +\beta }} d x, \alpha > 0, \beta > 0$

$\int_{0}^{\pi / 2} \frac{\sin^{2 \alpha -1} x \cos^{2 \beta -1} x}{\left(a^{2} \sin^{2} x+b^{2} \cos^{2} x\right)^{\alpha +\beta }} d x, \quad \alpha > 0, \quad \beta > 0, a > 0, \quad b > 0$;

$\int_{0}^{\pi / 2} \frac{\sin^{\alpha -1} x \cos^{\alpha -1} x}{\left(a^{2} \sin^{2} x+b^{2} \cos^{2} x\right)^{\alpha }} d x, \quad \alpha > 0, \quad a b > 0$;

$\int_{0}^{\pi / 2} \sin^{\alpha } x \cos^{\beta } x d x, \quad \alpha > -1, \quad \beta > -1$;

$\int_{0}^{\pi } \frac{\sin^{\alpha -1} x}{(1+\beta \cos x)^{\alpha }} d x, \quad 0 < \left|\beta \right| < 1, \quad \alpha > 0$;

$\int_{-1}^{1} \frac{(1+x)^{2 \alpha -1}(1-x)^{2 \beta -1}}{\left(1+x^{2}\right)^{\alpha +\beta }} d x, \quad \alpha > 0, \quad \beta > 0$;

$\int_{a}^{b} \frac{(x-a)^{\alpha }(b-x)^{\beta }}{(x+c)^{\alpha +\beta +2}} d x, \quad 0 < a < b, \quad c > 0, \quad \alpha > -1, \quad \beta > -1$.

$\int_{0}^{1} \frac{d x}{\sqrt{1-x^{4}}} = \frac{\Gamma^{2}(1 / 4)}{4 \sqrt{2 \pi }}$;

$\int_{0}^{\pi } \frac{d \theta }{\sqrt{3-\cos \theta }} = \frac{1}{4 \sqrt{\pi }} \Gamma^{2}\left(\frac{1}{4}\right)$ (Hint. Set $\cos \theta = 1-2 \sqrt{x})$;

$\int_{0}^{\pi }\left(\frac{\sin \theta }{1+r \cos \theta }\right)^{\alpha -1} \frac{d \theta }{1+r \cos \theta } = \frac{2^{\alpha -1}}{\left(1-r^{2}\right)^{\alpha / 2}} \frac{[\Gamma (\alpha / 2)]^{2}}{\Gamma (\alpha )},\left|r\right| < 1$

(Hint. Set $\operatorname {tg} \frac{\theta }{2} = \sqrt{\frac{1+r}{1-r}} \sqrt{x}$);

$\int_{0}^{1} \frac{d x}{\sqrt{1-x^{4}}} \int_{0}^{1} \frac{x^{2} d x}{\sqrt{1-x^{4}}} = \frac{\pi }{4}$;

$\int_{0}^{+\infty } e^{-x^{4}} d x \int_{0}^{+\infty } x^{2} e^{-x^{4}} d x = \frac{\pi }{8 \sqrt{2}}$;

$\prod_{k = 1}^{n} \int_{0}^{+\infty } x^{k-1} e^{-x^{n}} d x = \left(\frac{1}{n}\right)^{n-1 / 2}(2 \pi )^{(n-1) / 2}$

(Hint. Use the reflection formula (4) and the equality $\frac{z^{n}-1}{z-1} = \prod_{k = 1}^{n-1}\left(z-e^{2 k \pi i / n}\right)$.

$\int_{0}^{\infty } \int_{0}^{\infty }(x+y)^{p} e^{-x-y} d x d y, p > 0$

$\int_{0}^{\infty } \int_{0}^{\infty }\left(x^{2}+y^{2}\right)^{p} e^{-x^{2}-y^{2}} d x d y, p > -1$;

$\int_{0}^{\infty } \int_{0}^{\infty } \int_{0}^{\infty } e^{-\left(x^{2}+y^{2}+z^{2}\right)^{p}} d x d y d z, p > 0$

$\iint_{G}\left(x^{2}+y^{2}\right)^{p}\left(1-x^{2}-y^{2}\right)^{q} d x d y, p > -1, q > -1, G = \left\{ x > 0, y > 0, x^{2}+y^{2} < 1\right\}$;

$\iiint_{G} \frac{d x d y d z}{\left(t^{2}-x^{2}-y^{2}-z^{2}\right)^{p}}, p < 1, G = \left\{ x^{2}+y^{2}+z^{2} < t^{2}\right\}$.