Question

\( \int_{5\pi}^{25\pi} |\sin 2x + \cos 2x| \ dx = \)

• A. \( 20\sqrt{2} \)
• B. \( 10\sqrt{2} \)
• C. \( 40\sqrt{2} \)
• D. \( 80\sqrt{2} \)

Hint:

1. Rewrite \( A\sin\theta + B\cos\theta \) as \( R\sin(\theta+\alpha) \) or \( R\cos(\theta-\alpha) \), where \( R=\sqrt{A^2+B^2} \). 2. Use substitution to simplify the argument of the trigonometric function. 3. The function \( |\sin u| \) is periodic with period \( \pi \). 4. \( \int_0^\pi |\sin u| du = \int_0^\pi \sin u du = 2 \). 5. For a periodic function \(g(x)\) with period \(T\), \( \int_{a}^{a+nT} g(x)dx = n \int_0^T g(x)dx \).

Answer 1

The Correct Option is C

Let \( f(x) = \sin 2x + \cos 2x \).
We can write \( f(x) = \sqrt{1^2+1^2} \left( \frac{1}{\sqrt{2}}\sin 2x + \frac{1}{\sqrt{2}}\cos 2x \right) = \sqrt{2} (\cos(\pi/4)\sin 2x + \sin(\pi/4)\cos 2x) \).
So, \( f(x) = \sqrt{2} \sin(2x+\pi/4) \).
We need to evaluate \( I = \int_{5\pi}^{25\pi} |\sqrt{2} \sin(2x+\pi/4)| dx = \sqrt{2} \int_{5\pi}^{25\pi} |\sin(2x+\pi/4)| dx \).
Let \( u = 2x+\pi/4 \).
Then \( du = 2dx \implies dx = du/2 \).
When \( x=5\pi \), \( u = 10\pi+\pi/4 \).
When \( x=25\pi \), \( u = 50\pi+\pi/4 \).
\[ I = \sqrt{2} \int_{10\pi+\pi/4}^{50\pi+\pi/4} |\sin u| \frac{du}{2} = \frac{\sqrt{2}}{2} \int_{10\pi+\pi/4}^{50\pi+\pi/4} |\sin u| du \] The function \( |\sin u| \) is periodic with period \( \pi \).
The length of the interval of integration is \( (50\pi+\pi/4) - (10\pi+\pi/4) = 40\pi \).
This length is \( 40 \) times the period \( \pi \).
The integral of \( |\sin u| \) over one period \( [0, \pi] \) is \( \int_0^\pi \sin u du = [-\cos u]_0^\pi = (-\cos\pi) - (-\cos 0) = -(-1) - (-1) = 1+1=2 \).
So, \( \int_{10\pi+\pi/4}^{50\pi+\pi/4} |\sin u| du = 40 \times \int_0^\pi |\sin u| du \) (because the interval starts and ends at same phase relative to period boundary).
More precisely, \( \int_{a}^{a+nT} g(u)du = n \int_0^T g(u)du \) if \(g(u)\) has period T.
Here, the interval is \( [10\pi+\pi/4, (10\pi+\pi/4) + 40\pi] \).
Number of periods is 40.
So, \( \int_{10\pi+\pi/4}^{50\pi+\pi/4} |\sin u| du = 40 \times \int_0^\pi \sin u du = 40 \times 2 = 80 \).
Therefore, \( I = \frac{\sqrt{2}}{2} \times 80 = 40\sqrt{2} \).
This matches option (3).