For $U_n=\int_{0}^{1}x^n(2-x)^ndx,V_n=\int_{0}^{1}x^n(1-x)^ndx,n\in N$,which of the following statement is/are true
$(A)\ U_n=2^nV_n\hspace{1cm}(B)\ U_n=2^{-n}V_n\hspace{1cm}(C)\ U_n=2^{2n}V_n\hspace{1cm}(D)\ U_n=2^{-2n}V_n$
$V_n $does not change its form when $x$ is replaced by $1-x\Rightarrow$$V_n=\int_{0}^{1}x^n(1-x)^ndx$.
$U_n=\int_{0}^{1}(1-x)^n(1+x)^ndx$ .There is no way looking by which i can relate $U_n$ and $V_n$.
Given $$\displaystyle U_{n} = \int_{0}^{1}x^n\cdot (2-x)^ndx$$ and $$\displaystyle V_{n} = \int_{0}^{1}x^n\cdot (1-x)^ndx$$
Now Let $x=2y$ in $U_{n}\;,$ and $dx = 2dy$ and changing Limit, We get $\displaystyle $
$$\displaystyle U_{n}=2\int_{0}^{\frac{1}{2}}2^n\cdot y^n\cdot 2^n\cdot (1-y)^n dy $$
$$\displaystyle = 2^{n+n}\cdot 2\int_{0}^{\frac{1}{2}}y^n\cdot (1-y)^ndy = 2^{n+n}\cdot 2\int_{0}^{\frac{1}{2}}x^n\cdot (1-x)^ndx$$
Above we use the formula $$\displaystyle \int_{a}^{b}f(y)dy = \int_{a}^{b}f(t)dt$$
Now $$\displaystyle U_{n} = 2^{n+n}\cdot 2\int_{0}^{\frac{1}{2}}x^{n}\cdot (1-x)^ndx = 2^{n+n}\int_{0}^{1}x^n\cdot (1-x)^n = 2^{2n}\cdot V_{n}$$
So We get $$\displaystyle U_{n} = 2^{2n}\cdot V_{n}$$ above we use the formula $$2\displaystyle \int_{0}^{a}f(x)dx = \displaystyle \int_{0}^{2a}f(x)dx\;,$$ If $$f(2a-x) = f(x)$$
Like in above case $$\displaystyle 2\int_{0}^{\frac{1}{2}}x^n\cdot (1-x)^ndx = \int_{0}^{1}x^n\cdot (1-x)^ndx$$
Bcz here $$\displaystyle f(x)=x^n\cdot (1-x)^ndx.\;,$$ Then $$f(1-x) = (1-x)^n\cdot x^n = f(x)$$
Notice, $$\int_{0}^{1}x^n(2-x)^ndx$$ $$=\int_{0}^{1}(1-x)^n(2-(1-x))^ndx$$ $$=\int_{0}^{1}(1-x)^n(1+x)^ndx$$ $$=\int_{0}^{1}(1-x^2)^ndx$$ Now, let $x=\sin\theta\implies dx=\cos\theta d\theta$ $$\int_{0}^{\pi/2}(1-\sin^2\theta)^n\cos\theta d\theta$$ $$=\int_{0}^{\pi/2}cos^{2n}\theta\cos\theta d\theta$$ $$=\int_{0}^{\pi/2}cos^{2n+1}\theta d\theta$$ $$=\frac{\Gamma\left(n+1\right)\Gamma\left(\frac{1}{2}\right)}{2\Gamma\left(\frac{2n+3}{2}\right)}$$ $$=\frac{\Gamma\left(n+1\right)\sqrt {\pi}}{2\Gamma\left(\frac{2n+3}{2}\right)}$$