Question Detail

MTH 102 MTH 102 Past Questions Objective Question

Question

The Wallis formula for $\int_0^{\frac{\pi}{2}} \sin^n x dx$, where $n \geq 2$ and $n$ is an odd number, is
Options
A $\frac{(n-1)(n-3)(n-5)...6.4.2}{(n-2)(n-4)(n-6)...5.3.1}$
Correct Answer
B $\frac{4.(n-1)(n-3)(n-5)...6.4.2}{(n-2)(n-4)(n-6)...5.3.1} \frac{\pi}{2}$
C $\frac{4.(n-1)(n-3)(n-5)...6.4.2}{(n-2)(n-4)(n-6)...5.3.1}$
D $\frac{4.(n-1)(n-3)(n-5)...5.3.1}{(n-2)(n-4)(n-6)...6.4.2}$
Correct Answer

Option A is the correct answer.

Detailed Explanation

The standard Wallis formula for $\int_0^{\frac{\pi}{2}} \sin^n x dx$ when $n$ is an odd number is:$\frac{(n-1)!!}{n!!} = \frac{(n-1)(n-3)\dots 2}{n(n-2)\dots 1}$.Let's compare this with the given options. The options provided have a denominator that begins with $(n-2)$ instead of $n$. If we were to interpret option A as $\frac{(n-1)!!}{(n-2)!!}$, then for $n=3$, it gives $\frac{2}{1}=2$, and for $n=5$, it gives $\frac{4 \cdot 2}{3 \cdot 1} = \frac{8}{3}$. However, the standard formula yields $2/3$ for $n=3$ and $8/15$ for $n=5$.Given the provided options, and considering common variations or potential simplifications, option A represents a product of decreasing even terms in the numerator and decreasing odd terms in the denominator, albeit not in the standard $n!!$ form. Based on the marking in the source document (Page 10), option A is chosen as correct, despite the discrepancy with the exact standard formula for the denominator.

Hint

Recall the Wallis formula for definite integrals of powers of sine or cosine. Pay attention to whether $n$ is odd or even.

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