In I(m, n)=int 0 1 x m-1 (1-x) n-1 d x, m, n 0 , then I (9,14)+ I (10,13) is

Mathematics · Definite Integration

JEE Main 2025 — 24 January, Morning Shift — Question 2

In $I(m, n)=\int_{0}^{1} x^{m-1}(1-x)^{n-1} d x, m, n>0$ , then $\mathrm{I}(9,14)+\mathrm{I}(10,13)$ is

$I(m, m)=\int_{0}^{1} x^{m-1}(1-x)^{n-1} d x$

Let $x=\sin ^{2} \theta, \quad d x=2 \sin \theta \cos \theta d \theta$

$I(m, n)=2 \int_{0}^{\pi / 2}(\sin \theta)^{2 m-1}(\cos \theta)^{2 n-1} d \theta$

$I(9,14)+I(10,13)=2 \int_{0}^{\pi / 2}(\sin \theta)^{17}(\cos \theta)^{27} d \theta$

$+2 \int_{0}^{\pi / 2}(\sin \theta)^{19}(\cos \theta)^{25} \mathrm{~d} \theta$

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

$=I(9,13)$

Answer key and solution verified before publishing.

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