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 using a trigonometric identity.
We need $\int_0^{50\pi} \sqrt{1 - \cos 2x}\, dx$. Use the identity $1 - \cos 2x = 2\sin^2 x$: \[\sqrt{1 - \cos 2x} = \sqrt{2\sin^2 x} = \sqrt{2}|\sin x|\] The absolute value is necessary because $\sqrt{\sin^2 x} = |\sin x|$ is always non-negative.
Step 2: Identify the period of the integrand.
The function $|\sin x|$ has period $\pi$ (not $2\pi$), because $|\sin(x+\pi)| = |-\sin x| = |\sin x|$. So $\sqrt{2}|\sin x|$ has period $T = \pi$.
Step 3: Apply the periodic integral property.
The given property: $\int_0^{NT} f(t)\,dt = N\int_0^T f(t)\,dt$ with $T = \pi$ and $N = 50$: \[\int_0^{50\pi}\sqrt{2}|\sin x|\,dx = 50\int_0^\pi \sqrt{2}|\sin x|\,dx\]
Step 4: Compute the integral over one period [0, pi].
On $[0, \pi]$, $\sin x \geq 0$, so $|\sin x| = \sin x$: \[\sqrt{2}\int_0^\pi \sin x\,dx = \sqrt{2}[-\cos x]_0^\pi = \sqrt{2}(1+1) = 2\sqrt{2}\]
Step 5: Multiply by the number of periods.
\[\int_0^{50\pi}\sqrt{1-\cos 2x}\,dx = 50 \times 2\sqrt{2} = 100\sqrt{2}\]
Step 6: State the final answer.
\[\boxed{100\sqrt{2}}\]