If $$n(2n+1) \int_{0}^{1}\left(1-x^{n}\right)^{2n} dx=1177 \int_{0}^{1}\left(1-x^{n}\right)^{2n+1} dx,$$ then $n \in \mathbb{N}$ is equal to ______.

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Definite Integrals

n (2 n + 1) \int_{0}^{1} (1 - x^{n})^{2 n} d x = 1177 \int_{0}^{1} (1 - x^{n})^{2 n + 1} d x,

then

n \in N

is equal to ______.

Answer: 24

Key concept: Integration by parts and algebraic manipulation.

Let

I_{1} = \int_{0}^{1} (1 - x^{n})^{2 n} d x

and

I_{2} = \int_{0}^{1} (1 - x^{n})^{2 n + 1} d x

.

Applying integration by parts on

I_{2}

I_{2} = [(1 - x^{n})^{2 n + 1} \cdot x]_{0}^{1} + n (2 n + 1) \int_{0}^{1} (1 - x^{n})^{2 n} x^{n} d x

I_{2} = n (2 n + 1) (I_{1} - I_{2})

(2 n^{2} + n + 1) I_{2} = n (2 n + 1) I_{1}

From the given condition:

\frac{I _{1}}{I _{2}} = \frac{1177}{n ( 2 n + 1 )} = \frac{2 n ^{2} + n + 1}{n ( 2 n + 1 )}

Therefore:

2 n^{2} + n + 1 = 1177

2 n^{2} + n - 1176 = 0

Solving:

n = 24

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