Question

The value of the definite integral $\int_{0}^{\frac{\pi}{2}} (\cos^{10}x \cdot \sin 12x) dx$, is equal to

• A. 1/10
• B. 1/11
• C. 1/12
• D. 1/22

Answer 1

Answer: (B)

1/11

Answer 2

The problem asks for the value of the definite integral $I = \int_{0}^{\frac{\pi}{2}} (\cos^{10}x \cdot \sin 12x) dx$. The integral is of the form $\int_{0}^{\frac{\pi}{2}} \cos^m x \sin^n x \, dx$. For this type of integral, the Wallis' integral formula states: $\int_{0}^{\frac{\pi}{2}} \cos^m x \sin^n x \, dx = \begin{cases} \frac{(m-1)!! (n-1)!!}{(m+n)!!} & \text{if } m+n \text{ is odd} \\ \frac{(m-1)!! (n-1)!!}{(m+n)!!} \frac{\pi}{2} & \text{if } m \text{ and } n \text{ are both even} \end{cases}$ In the given problem, $m=10$ and $n=12$. Both $m$ and $n$ are even, and $m+n = 22$ is also even. According to the formula, the integral should be: $I = \frac{(10-1)!! (12-1)!!}{(10+12)!!} \frac{\pi}{2} = \frac{9!! \cdot 11!!}{22!!} \frac{\pi}{2}$ This result involves $\pi$, but the given options are simple fractions without $\pi$. This strongly suggests a typo in the question. A common scenario is that one of the powers was intended to be 1, leading to an odd sum of powers.

Let's assume there was a typo and the integral was intended to be $\int_{0}^{\frac{\pi}{2}} (\cos^{10}x \cdot \sin^1 x) dx$. In this case, $m=10$ and $n=1$. The sum of the powers is $m+n = 10+1 = 11$, which is odd. Using the Wallis' integral formula for an odd sum of powers: $I = \frac{(m-1)!! (n-1)!!}{(m+n)!!}$ Substituting $m=10$ and $n=1$: $I = \frac{(10-1)!! (1-1)!!}{(10+1)!!} = \frac{9!! \cdot 0!!}{11!!}$ By convention, $0!! = 1$. $I = \frac{9!!}{11!!}$ The double factorial $k!!$ is the product of all integers from $k$ down to 1 with the same parity as $k$. $9!! = 9 \times 7 \times 5 \times 3 \times 1$ $11!! = 11 \times 9 \times 7 \times 5 \times 3 \times 1$ Therefore, $I = \frac{9 \times 7 \times 5 \times 3 \times 1}{11 \times 9 \times 7 \times 5 \times 3 \times 1} = \frac{1}{11}$ This result matches option (b). Given the discrepancy between the calculated value for the original question (which includes $\pi$) and the provided options, it is reasonable to conclude that the question likely contained a typo and was intended to be $\int_{0}^{\frac{\pi}{2}} (\cos^{10}x \cdot \sin^1 x) dx$, yielding the answer $\frac{1}{11}$.