First slide

Evaluation of definite integrals

Question

The value of $lim_{m \to \infty} \frac{\int_{0}^{π / 2} \sin^{2 m} x d x}{\int_{0}^{π / 2} \sin^{2 m + 1} x d x}$

Moderate

A

0
B

1/2
C

2
D

1

Solution

We know that $I_{2 n} = \int_{0}^{π / 2} \sin^{2 n} x d x$
$\begin{array}{l} = \frac{2 n - 1}{2 n} \times \frac{2 n - 3}{2 n - 2} \times \dots \times \frac{1}{2} \times \frac{π}{2}, \\ I_{2 n + 1} = \int_{0}^{π / 2} \sin^{2 n + 1} x d x \end{array}$
$= \frac{2 n}{2 n + 1} \times \frac{2 n - 2}{2 n - 1} \times \dots \times \frac{2}{3}$ and
Also, $I_{2 m + 1} = \frac{2 m}{2 m + 1} I_{2 m - 1}$
For all $x \in (0, π / 2), \sin^{2 m - 1} x > \sin^{2 m} x > \sin^{2 m + 1} x$
Integrating from 0 to $π / 2$ , we get $I_{2 m - 1} \geq I_{2 m} \geq I_{2 m + 1}$
hence $\frac{I_{2 m - 1}}{I_{2 m + 1}} \geq \frac{I_{2 m}}{I_{2 m + 1}} \geq 1$ (i)
Also $\frac{I_{2 m - 1}}{I_{2 m + 1}} = \frac{2 m + 1}{2 m}$
Hence $lim_{m \to \infty} \frac{I_{2 m - 1}}{I_{2 m + 1}} = lim_{m \to \infty} \frac{2 m + 1}{2 m} = 1$
From (i) and using sandwitch theorem we have
$lim_{m \to \infty} \frac{I_{2 m}}{I_{2 m + 1}} = 1$

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