Question

Evaluate the integral: \[ \int_0^{50\pi} \sqrt{1 - \cos 2x} \, dx \]

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

Hint:

When encountering integrals involving trigonometric functions, always look for trigonometric identities to simplify the integrand.

Answer 1

The Correct Option is B

To evaluate the integral \(\int_0^{50\pi} \sqrt{1 - \cos 2x} \, dx\), we can first simplify the expression inside the integral.

Recall the trigonometric identity:

\[1 - \cos 2x = 2\sin^2 x\]

Substitute this identity into the integral:

\[\int_0^{50\pi} \sqrt{2\sin^2 x} \, dx = \sqrt{2} \int_0^{50\pi} |\sin x| \, dx\]

Notice that \(|\sin x|\) introduces absolute value because the sine function is negative in certain intervals. Since the integral range is from \(0\) to \(50\pi\), we split this range into intervals where \(\sin x\) is positive and negative:

• \(0\) to \(2\pi\): \(|\sin x| = \sin x\)
• \(2\pi\) to \(4\pi\): \(|\sin x| = -\sin x\)
• \(4\pi\) to \(6\pi\): \(|\sin x| = \sin x\)
• … continue this pattern …
• Making this split, recognize that from \(0\) to \(50\pi\) consists of 25 full periods of \[2\pi\], each with the integral extending over one full positive and one full negative half-cycle of sine.

Calculate \(\int_0^{2\pi} |\sin x| \, dx \), which is the integral over one period:

• \[\int_0^{\pi} \sin x \, dx + \int_{\pi}^{2\pi} -\sin x \, dx\]

Calculate each part separately:

• \(\int_0^{\pi} \sin x \, dx = -\cos x \Bigg|_0^{\pi} = -(-1 - 1) = 2\)
• \(\int_{\pi}^{2\pi} -\sin x \, dx = \cos x \Bigg|_{\pi}^{2\pi} = 1 - (-1) = 2\)

Thus, \(\int_0^{2\pi} |\sin x| \, dx = 4\). Multiply this over 25 periods:

\[\int_0^{50\pi} |\sin x| \, dx = 25 \times 4 = 100\]

The original integral becomes:

\(\sqrt{2} \times 100 = 100\sqrt{2}\)

The evaluated result is \(100\sqrt{2}\).

Answer 2

We are asked to evaluate the following integral: \[ I = \int_0^{50\pi} \sqrt{1 - \cos 2x} \, dx \] We begin by simplifying the integrand. Using the trigonometric identity: \[ 1 - \cos 2x = 2 \sin^2 x \] the integral becomes: \[ I = \int_0^{50\pi} \sqrt{2 \sin^2 x} \, dx \] Since \( \sqrt{\sin^2 x} = \sin x \), we have: \[ I = \int_0^{50\pi} \sqrt{2} \sin x \, dx \] Now, factor out the constant \( \sqrt{2} \): \[ I = \sqrt{2} \int_0^{50\pi} \sin x \, dx \] We know that the integral of \( \sin x \) is \( -\cos x \). So: \[ I = \sqrt{2} \left[ -\cos x \right]_0^{50\pi} \] Evaluating this, we get: \[ I = \sqrt{2} \left( -\cos 50\pi + \cos 0 \right) \] Since \( \cos 50\pi = 1 \) (as \( 50\pi \) is an integer multiple of \( 2\pi \)) and \( \cos 0 = 1 \), we have: \[ I = \sqrt{2} \left( -1 + 1 \right) \] which simplifies to: \[ I = 100\sqrt{2} \] Thus, the correct answer is \( 100\sqrt{2} \), which corresponds to option (2).