Walli's Formula
Evaluate the definite integral from zero to \pi/2 of cosine raised to the fourth power, applying the reduction relationship encapsulated by Walli's formula for even powers.
Select the correct option:
Solution
3π/16
Powers of sine or cosine integrated over the quarter period [0, \pi/2] are governed by Walli's formula, which compresses an otherwise repetitive integration-by-parts recursion into a single ratio. For an even exponent n the formula reads \int_0^{\pi/2}\cos^n x,dx = \frac{(n-1)(n-3)\cdots 1}{n(n-2)\cdots 2}\cdot\frac{\pi}{2}, where the extra \pi/2 factor appears only for even powers. For n = 4 this gives \frac{3\cdot 1}{4\cdot 2}\cdot\frac{\pi}{2} = \frac{3}{8}\cdot\frac{\pi}{2} = \frac{3\pi}{16}. As an independent confirmation, writing \cos^4 x = \left(\frac{1+\cos 2x}{2}\right)^2 and expanding into constant and double-angle terms reproduces the same value after integration. Option \pi/4 corresponds to \cos^2 x, which is one power too few. Option \pi/2 is the integral of 1, ignoring the fourth power entirely. Option 3\pi/8 omits the final \pi/2 factor that even powers demand. The formula neatly systematizes this whole family of symmetric power integrals. As a final plausibility check, since \cos^4 x \le 1 everywhere on the interval, the integral must stay strictly below the integral of one over the same range, which is \pi/2 \approx 1.571, and the computed value 3\pi/16 \approx 0.589 sits comfortably within that upper bound, exactly as a fourth power averaging well under one would predict.
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About This Question
- Subject
- mathematics
- Chapter
- integral calculus
- Topic
- walli's formula
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
3π/16
Powers of sine or cosine integrated over the quarter period [0, \pi/2] are governed by Walli's formula, which compresses an otherwise repetitive integration-by-parts recursion into a single ratio. For an even exponent n the formula reads \int_0^{\pi/2}\cos^n x,dx = \frac{(n-1)(n-3)\cdots 1}{n(n-2)\cdots 2}\cdot\frac{\pi}{2}, where the extra \pi/2 factor appears only for even powers. For n = 4 this gives \frac{3\cdot 1}{4\cdot 2}\cdot\frac{\pi}{2} = \frac{3}{8}\cdot\frac{\pi}{2} = \frac{3\pi}{16}. As an independent confirmation, writing \cos^4 x = \left(\frac{1+\cos 2x}{2}\right)^2 and expanding into constant and double-angle terms reproduces the same value after integration. Option \pi/4 corresponds to \cos^2 x, which is one power too few. Option \pi/2 is the integral of 1, ignoring the fourth power entirely. Option 3\pi/8 omits the final \pi/2 factor that even powers demand. The formula neatly systematizes this whole family of symmetric power integrals. As a final plausibility check, since \cos^4 x \le 1 everywhere on the interval, the integral must stay strictly below the integral of one over the same range, which is \pi/2 \approx 1.571, and the computed value 3\pi/16 \approx 0.589 sits comfortably within that upper bound, exactly as a fourth power averaging well under one would predict.
This medium difficulty mathematics question is from the chapter integral calculus, covering the topic of walli's formula. It appeared in the 2025 exam.
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