Question

If \(f\) is a continuous function and \(f(x+T)=f(x), \forall x\in R\), it is given that \[ \int_0^{NT} f(t)\,dt=N\int_0^T f(t)\,dt \] \((N\) is a natural number). Then \[ \int_0^{50\pi}\sqrt{1-\cos 2x}\,dx= \]

• A. \(50\sqrt{2}\)
• B. \(100\sqrt{2}\)
• C. \(\frac{50}{\sqrt{2}}\)
• D. \(\frac{100}{\sqrt{2}}\)

Hint:

Use \(1-\cos 2x=2\sin^2x\), but remember that \(\sqrt{\sin^2x}=|\sin x|\), not simply \(\sin x\).

Answer 1

The Correct Option is B

Step 1: Simplify the integrand.
We know that \[ 1-\cos 2x=2\sin^2x \] Therefore, \[ \sqrt{1-\cos 2x} = \sqrt{2\sin^2x} \] \[ = \sqrt{2}|\sin x| \]

Step 2: Use periodicity.
The function \[ |\sin x| \] has period \[ \pi \] Therefore, \[ \sqrt{2}|\sin x| \] also has period \[ \pi \]

Step 3: Integrate over one period.
Over one period, \[ \int_0^\pi \sqrt{2}|\sin x|\,dx = \sqrt{2}\int_0^\pi \sin x\,dx \] \[ = \sqrt{2}\left[-\cos x\right]_0^\pi \] \[ = \sqrt{2}\left[1+1\right] \] \[ = 2\sqrt{2} \]

Step 4: Use the given periodic integral property.
Since \[ 50\pi=50\times \pi, \] we get \[ \int_0^{50\pi}\sqrt{1-\cos 2x}\,dx = 50\int_0^\pi \sqrt{2}|\sin x|\,dx \] \[ = 50(2\sqrt{2}) \] \[ = 100\sqrt{2} \]

Step 5: Final conclusion.
Hence, \[ \boxed{100\sqrt{2}} \]