If n 2 n + 1 ∫ 0 1 1 - x n 2 n d x = 1177 ∫ 0 1 1 - x n 2 n + 1 d x , then n ∈ N is equal…

Let $I_1={\int }_0^1{\left(1-x^n\right)}^{2n}dx$ and $I_2={\int }_0^1{\left(1-x^n\right)}^{2n+1}dx$

Now on solving $I_2$ we get,

$I_2={\int }_0^1{\left(1-x^n\right)}^{2n+1}·1dx$

Now on using by parts integration we get,

$\Rightarrow I_2={\left(1-x^n\right)}^{2n+1}.{\overline{)x}}_0^1-{\int }_0^1\left(2n+1\right){\left(1-x^n\right)}^{2n}\left(-n{x}^{n-1}\right)xdx$

$\Rightarrow I_2=-n\left(2n+1\right)\left\{I_2-I_1\right\}$ $\left\{\text{as }{I}_1={\int }_0^1{\left(1-x^n\right)}^{2n}dx\right\}$

$\Rightarrow \left(2n^2+n+1\right)I_2=n\left(2n+1\right)I_1$

$\left\{\text{Given }n\left(2n+1\right){\int }_0^1{\left(1-x^n\right)}^{2n}dx=1177{\int }_0^1{\left(1-x^n\right)}^{2n+1}dx\right\}$

$\Rightarrow \frac{I_1}{I_2}=\frac{2n^2+n+1}{n\left(2n+1\right)}=\frac{1177}{n\left(2n+1\right)}$

$\Rightarrow 2n^2+n-1176=0$

$\Rightarrow n=\frac{-1\pm \sqrt{1+4\times 2\times 1176}}{4}$

$\Rightarrow n=\frac{-1\pm \sqrt{9409}}{4}$

$\Rightarrow n=\frac{-1\pm 97}{4}$

$\Rightarrow n=24 or \frac{-49}{2}$

But given $n\in N$ so $n=24$