Question

The integral \( \int_{0}^{\pi/2} \sin^5 x \cos^7 x \,dx = \)

• A. \(\pi\)
• B. 120
• C. \(\frac{1}{120}\)
• D. 1

Hint:

The reduction formula is a powerful shortcut for integrals of \(\sin^m x \cos^n x\) from 0 to \(\pi/2\). Just remember to start decrementing by 2 from \(m-1\), \(n-1\) in the numerator and from \(m+n\) in the denominator. And don't forget the \(\pi/2\) factor if both powers are even!

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This integral is a specific case of Wallis' integrals, which have a standard reduction formula, especially for definite integrals from 0 to \(\pi/2\). This is related to the Beta function.
Step 2: Key Formula or Approach:
The reduction formula for integrals of the form \( \int_{0}^{\pi/2} \sin^m x \cos^n x \,dx \) is given by: \[ \int_{0}^{\pi/2} \sin^m x \cos^n x \,dx = \frac{[(m-1)(m-3)\dots][(n-1)(n-3)\dots]}{(m+n)(m+n-2)\dots} \times K \] where the terms in the numerator continue until they reach 1 or 2.
The value of \(K\) is:
- \(K = \frac{\pi}{2}\) if both \(m\) and \(n\) are even.
- \(K = 1\) otherwise (if at least one of \(m\) or \(n\) is odd).
Step 3: Detailed Explanation:
Using the reduction formula for \(m=5\) and \(n=7\).
Since at least one power is odd (in this case, both are), the factor \(K\) will be 1.
Numerator:
\((m-1)(m-3)\dots = (5-1)(5-3) = 4 \times 2\)
\((n-1)(n-3)\dots = (7-1)(7-3)(7-5) = 6 \times 4 \times 2\)
Denominator:
\((m+n)(m+n-2)\dots = (5+7)(5+7-2)(5+7-4)(5+7-6)(5+7-8)(5+7-10) = 12 \times 10 \times 8 \times 6 \times 4 \times 2\)
Now, assemble the fraction: \[ \int_{0}^{\pi/2} \sin^5 x \cos^7 x \,dx = \frac{(4 \times 2) \times (6 \times 4 \times 2)}{12 \times 10 \times 8 \times 6 \times 4 \times 2} \] We can cancel the term \((6 \times 4 \times 2)\) from the numerator and the denominator: \[ = \frac{4 \times 2}{12 \times 10 \times 8} = \frac{8}{960} \] Simplify the fraction: \[ = \frac{1}{120} \] Step 4: Final Answer:
The value of the integral is \(\frac{1}{120}\). This matches option (3).