Could someone help me out with calculating this integral:
$\int_{0}^{u} 1/(\sqrt{1-x} \sqrt{x-t})\, dx$
I tried integration by part (maybe not far enough) and it did not seem to bring me to a solution. I also used square completion without success.
Could someone help me out with calculating this integral:
$\int_{0}^{u} 1/(\sqrt{1-x} \sqrt{x-t})\, dx$
I tried integration by part (maybe not far enough) and it did not seem to bring me to a solution. I also used square completion without success.
Completing the square should work.
$$\int_{0}^{u} \frac{1}{\sqrt{1-x} \sqrt{x-t}}=\int_{0}^{u} \frac{1}{ \sqrt{-x^2+(t+1)x-1}})\, dx =$$ $$=\int_{0}^{u} \frac{1}{ \sqrt{-x^2+(t+1)x-\frac{t^2+2t+1}{4}+\frac{t^2+2t-3}{4}}}\, dx $$
Now, if $t^2+2t-3 <0$ the function is not well defined. Otherwise, setting $\alpha=\sqrt{\frac{t^2+2t-3}{4}}$ you have
$$\int_0^u \frac{1}{\sqrt{\alpha^2-(x-\frac{t+1}{2})^2}}$$
which can be calculated with the substitution $x-\frac{t+1}{2}=\alpha \sin (v)$. Of course you need to put some extra restrictions on $t$, to make sure that your function is defined on $[0,u]$.
Try some changes of variable: first $x = 1 - u^2$, then $u = \sqrt{1-t}\; v$...