Evaluation of $$ 2^{2010}\frac{\int^{1}_{0}x^{1004}(1-x)^{1004}dx}{\int^{1}_{0}x^{1004}(1-x^{2010})^{1004}dx}$$
Try: Let $\displaystyle I =\int^{1}_{0}x^{1004}(1-x)^{1004}dx$ and $\displaystyle J =\int^{1}_{0}x^{1004}(1-x^{2010})^{1004}dx$
put $x^{1005}=t$ and $\displaystyle x^{1004}dx=\frac{1}{1005}dt$
So $\displaystyle J =\frac{1}{1005}\int^{1}_{0}(1-t^2)dt$
Now i am struck here , I did not understand how to solve it
Could some help me to solve it, Thanks
Best Answer
With the substitution $x^{1005} = t$, we have $$J = \frac{1}{1005}\int_0^1 (1 - t^2)^{1004} \mathrm{d} t. \tag{1}$$
With the substitution $t = 1 - 2u$, we have $$J = \frac{1}{1005}\int_0^{1/2} 2^{2009}u^{1004}(1 - u)^{1004}\mathrm{d} u . \tag{2}$$
With the substitution $u = 1 - v$, we have $$J = \frac{1}{1005}\int_{1/2}^1 2^{2009}v^{1004}(1 - v)^{1004}\mathrm{d} v. \tag{3}$$
Using (2) and (3), we have $$J = \frac12\cdot \frac{1}{1005} \int_0^1 2^{2009}u^{1004}(1 - u)^{1004}\mathrm{d} u = \frac{1}{4020}2^{2010}I.$$
Thus, $$2^{2010}I/J = 4020.$$