If n(2n + 1) ∫ _ 0 ^ 1 (1 - x ^ n ) ^ 2n dx = 1177 ∫ _ 0 ^ 1 (1 - x...

Let ${I}_{1} = {∫}_{0}^{1}{\left(1 - {x}^{n}\right)}^{2n}dx$ and ${I}_{2} = {∫}_{0}^{1}{\left(1 - {x}^{n}\right)}^{2n + 1}dx$

Now on solving ${I}_{2}$ we get,

${I}_{2} = {∫}_{0}^{1}{\left(1 - {x}^{n}\right)}^{2n + 1} \cdot 1dx$

Now on using by parts integration we get,

$⇒ {I}_{2} = {\left(1 - {x}^{n}\right)}^{2n + 1} . {x}_{0}^{1} - {∫}_{0}^{1}\left(2n + 1\right){\left(1 - {x}^{n}\right)}^{2n}\left(- n{x}^{n - 1}\right)xdx$

$⇒ {I}_{2} = - n\left(2n + 1\right)\left\{{I}_{2} - {I}_{1}\right\} \left\{\text{as}{I}_{1} = {∫}_{0}^{1}{\left(1 - {x}^{n}\right)}^{2n}dx\right\}$

$⇒ \left(2{n}^{2} + n + 1\right){I}_{2} = n\left(2n + 1\right){I}_{1}$

\left\{\text{Given}n\left(2n + 1\right){∫}_{0}^{1}{\left(1 - {x}^{n}\right)}^{2n}dx = 1177{∫}_{0}^{1}{\left(1 - {x}^{n}\right)}^{2n + 1}dx\right\}$</p><p>$⇒ \dfrac{{I}_{1}}{{I}_{2}} = \dfrac{2{n}^{2} + n + 1}{n\left(2n + 1\right)} = \dfrac{1177}{n\left(2n + 1\right)}$</p><p>$⇒ 2{n}^{2} + n - 1176 = 0$</p><p>$⇒ n = \dfrac{- 1 \pm \sqrt{1 + 4 \times 2 \times 1176}}{4}$</p><p>$⇒ n = \dfrac{- 1 \pm \sqrt{9409}}{4}$</p><p>$⇒ n = \dfrac{- 1 \pm 97}{4}$</p><p>$⇒ n = 24or\dfrac{- 49}{2}$</p><p>But given$n \in N$so$n = 24