Question

The value of \( \int_0^{20} (\sin 4x + \cos 4x) \, dx \) is equal to:

• A. \( \frac{15\pi}{2} \)
• B. \( \frac{25\pi}{2} \)
• C. \( 20\pi \)
• D. \( \frac{5\pi}{2} \)

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The integrand \(f(x) = \sin^4 x + \cos^4 x\) is a periodic function.
We need to determine its fundamental period and use the property of definite integrals for periodic functions: \(\int_0^{nT} f(x) dx = n \int_0^T f(x) dx\).
Step 2: Key Formula or Approach:
The period of both \(\sin^4 x\) and \(\cos^4 x\) is \(\pi\), but their sum actually has a shorter period.
Observe that \(f(x + \pi/2) = \sin^4(x + \pi/2) + \cos^4(x + \pi/2) = \cos^4 x + \sin^4 x = f(x)\).
Thus, the fundamental period \(T\) is \(\frac{\pi}{2}\).
Wallis' formula for integrals of the form \(\int_0^{\pi/2} \sin^n x dx = \int_0^{\pi/2} \cos^n x dx = \frac{(n-1)(n-3)\dots}{n(n-2)\dots} \cdot \frac{\pi}{2}\) (for even \(n\)).
Step 3: Detailed Explanation:
Since the period is \(T = \frac{\pi}{2}\), we can rewrite the upper limit \(20\pi\) as \(40 \times \frac{\pi}{2}\).
\[ I = \int_0^{40 \cdot \frac{\pi}{2}} (\sin^4 x + \cos^4 x) dx = 40 \int_0^{\pi/2} (\sin^4 x + \cos^4 x) dx \] Using Wallis' formula for \(n=4\), we evaluate the integrals of \(\sin^4 x\) and \(\cos^4 x\) separately.
\[ \int_0^{\pi/2} \sin^4 x dx = \frac{3 \cdot 1}{4 \cdot 2} \cdot \frac{\pi}{2} = \frac{3\pi}{16} \] Similarly, by symmetry:
\[ \int_0^{\pi/2} \cos^4 x dx = \frac{3\pi}{16} \] Now, substitute these back into the total integral.
\[ I = 40 \left( \frac{3\pi}{16} + \frac{3\pi}{16} \right) = 40 \left( \frac{6\pi}{16} \right) \] Simplify the fraction.
\[ I = 40 \left( \frac{3\pi}{8} \right) = 5 \cdot 3\pi = 15\pi \] Step 4: Final Answer:
The value of the integral is \(15\pi\).