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: Concept Overview:
This integral is a Wallis' integral, solvable with a standard reduction formula, particularly for definite integrals from 0 to \(\pi/2\). It connects to the Beta function.
Step 2: Key Formula:
The reduction formula for \( \int_{0}^{\pi/2} \sin^m x \cos^n x \,dx \) is: \[ \int_{0}^{\pi/2} \sin^m x \cos^n x \,dx = \frac{[(m-1)(m-3)\dots][(n-1)(n-3)\dots]}{(m+n)(m+n-2)\dots} \times K \]where numerator terms continue to 1 or 2.
\(K\) is:
- \(K = \frac{\pi}{2}\) if \(m\) and \(n\) are even.
- \(K = 1\) otherwise.
Step 3: Detailed Calculation:
Applying the formula with \(m=5\) and \(n=7\):
Since at least one power is odd (\(m=5\) and \(n=7\)), \(K = 1\).
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:
\((m+n)(m+n-2)\dots = (5+7)(5+7-2)(5+7-4)(5+7-6)(5+7-8)(5+7-10) = 12 \times 10 \times 8 \times 6 \times 4 \times 2\)
The integral becomes: \[ \int_{0}^{\pi/2} \sin^5 x \cos^7 x \,dx = \frac{(4 \times 2) \times (6 \times 4 \times 2)}{12 \times 10 \times 8 \times 6 \times 4 \times 2} \]Cancel \((6 \times 4 \times 2)\): \[ = \frac{4 \times 2}{12 \times 10 \times 8} = \frac{8}{960} \]Simplify: \[ = \frac{1}{120} \] Step 4: Answer:
The integral's value is \(\frac{1}{120}\), matching option (3).