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Problem 74 Hard Difficulty
Let $ I_n = \displaystyle \int_0^{\frac{\pi}{2}} \sin^n x dx $.
(a) Show that $ I_{2n + 2} \le I_{2n + 1} \le I_{2n} $.
(b) Use Exercise 50 to show that
$$ \frac{I_{2n + 2}}{I_{2n}} = \frac{2n + 1}{2n + 2} $$
(c) Use parts (a) and (b) to show that
$$ \frac{2n + 1}{2n + 2} \le \frac{I_{2n + 1}}{I_{2n}} \le 1 $$
and deduce that $ \lim_{n\to\infty}\frac{I_{2n + 1}}{I_{2n}} = 1 $.
(d) Use part (c) and Exercises 49 and 50 to show that
$$ \lim_{n\to\infty} \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdots \cdots \frac{2n}{2n - 1} \cdot \frac{2n}{2n + 1} = \frac{\pi}{2} $$
This formula is usually written as an infinite product:
$$ \frac{\pi}{2} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdots $$
and is called the Wallis product.
(e) We construct rectangles as follows. Start with a square of area 1 and attach rectangles of area 1 alternately beside or on top of the previous rectangle (see the figure). Find the limit of the ratios of width to height of these rectangles.
Answer
a) $\int_{0}^{\pi / 2} \sin ^{2 n+2} x d x \leq \int_{0}^{\pi / 2} \sin ^{2 n+1} x d x \leq \int_{0}^{\pi / 2} \sin ^{2 n} x d x$
So $, I_{2 n+2} \leq I_{2 n+1} \leq I_{2 n}$
b) Figure out the expression for $I_{2 n+2} .$ The part of $I_{2 n+2}$ that is the same as
$I_{2 n}$ cancels out easily.
c) 1
d) $\lim _{n \rightarrow \infty} \frac{1}{(2 n+1)}\left(\frac{2^{n}(n !)}{(2 n) !}\right)^{2}\left(2^{n}(n !)\right)^{2}=1$
$\lim _{n \rightarrow \infty} \frac{1}{(2 n+1)}\left(\frac{2^{n}(n !)}{(2 n) !}\right)^{2}\left(2^{n}(n !)\right)^{2}=\frac{\pi}{2}$
e) $\frac{\pi}{2}$
Topics
Integration Techniques
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Video Transcript
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Topics
Integration Techniques
Top Calculus 2 / BC Educators
Heather Z.
Oregon State University
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Harvey Mudd College
Caleb E.
Baylor University
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Boston College
