Definite Integration – Reduction formula – JEE Main 24 January 2025 Shift 1

Solution:

Given the integral:
$$ I(m,n) = \int_{0}^{1} x^{m-1}(1-x)^{n-1} dx $$

Let $x = \sin^2\theta$, then $dx = 2\sin\theta\cos\theta d\theta$.

The limits change: when $x=0, \theta=0$ and when $x=1, \theta=\frac{\pi}{2}$.

Substituting these into the integral:
$$ I(m,n) = \int_{0}^{\pi/2} (\sin^2\theta)^{m-1} (1-\sin^2\theta)^{n-1} (2\sin\theta\cos\theta) d\theta $$
$$ I(m,n) = 2 \int_{0}^{\pi/2} (\sin\theta)^{2m-2} (\cos^2\theta)^{n-1} (\sin\theta\cos\theta) d\theta $$
$$ I(m,n) = 2 \int_{0}^{\pi/2} (\sin\theta)^{2m-1} (\cos\theta)^{2n-1} d\theta $$

Now, find the values for the required terms:
For $I(9,14)$: $m=9, n=14$
$$ I(9,14) = 2 \int_{0}^{\pi/2} (\sin\theta)^{2(9)-1} (\cos\theta)^{2(14)-1} d\theta = 2 \int_{0}^{\pi/2} (\sin\theta)^{17} (\cos\theta)^{27} d\theta $$

For $I(10,13)$: $m=10, n=13$
$$ I(10,13) = 2 \int_{0}^{\pi/2} (\sin\theta)^{2(10)-1} (\cos\theta)^{2(13)-1} d\theta = 2 \int_{0}^{\pi/2} (\sin\theta)^{19} (\cos\theta)^{25} d\theta $$

Add the two integrals:
$$ I(9,14) + I(10,13) = 2 \int_{0}^{\pi/2} [ (\sin\theta)^{17} (\cos\theta)^{27} + (\sin\theta)^{19} (\cos\theta)^{25} ] d\theta $$

Take common factors $(\sin\theta)^{17} (\cos\theta)^{25}$ out:
$$ = 2 \int_{0}^{\pi/2} (\sin\theta)^{17} (\cos\theta)^{25} [ \cos^2\theta + \sin^2\theta ] d\theta $$

Since $\cos^2\theta + \sin^2\theta = 1$:
$$ = 2 \int_{0}^{\pi/2} (\sin\theta)^{17} (\cos\theta)^{25} d\theta $$

We need to express this back in the form $I(m,n)$. Comparing powers:
$$ 2m-1 = 17 \Rightarrow 2m=18 \Rightarrow m=9 $$
$$ 2n-1 = 25 \Rightarrow 2n=26 \Rightarrow n=13 $$

Thus, the expression is equivalent to $I(9,13)$.

Ans. (4)