Question

The function \(f(x) = \int _0^x\frac{dt}{1+cost}\) satisfies which of the following differential equations?

• A. \(2\frac{df}{dx} = 1+[f(x)]^2\)
• B. \(\frac{df}{dx} = 1+[f(x)]^2\)
• C. \(2\frac{df}{dx} = 1-[f(x)]^2\)
• D. \(\frac{df}{dx} = 1-[f(x)]^2\)

Hint:

Split the interval where sin 2x changes sign and use symmetry.

Answer 1

The Correct Option is A

Step 1: Sign of sin 2x:
\(\sin2x \ge 0\) for \(0 \le x \le \frac\pi2\) and \(\sin2x \le 0\) for \(\frac\pi2 \le x \le \pi\).

Step 2: Split the integral:
\[ \int_0^\pi|\sin2x|\,dx = \int_0^{\pi/2}\sin2x\,dx - \int_{\pi/2}^{\pi}\sin2x\,dx \]
\[ \int_0^{\pi/2}\sin2x\,dx = \left[-\frac{\cos2x}{2}\right]_0^{\pi/2} = \frac12 + \frac12 = 1 \]
\[ -\int_{\pi/2}^\pi\sin2x\,dx = \left[\frac{\cos2x}{2}\right]_{\pi/2}^{\pi} = \frac12 + \frac12 = 1 \]

Step 3: Total:
The sum is \(1 + 1 = 2\). Without the modulus, the integral would be 0 (option A), since the two halves cancel.

Final Answer:
The value is 2, option (C). \[ \boxed{2} \]