Question

Let f be a non-negative function defined on [\(0,\frac{\pi}{2}\)]. If \(\int_{0}^{x}(f'(t)-sin\,2t)dt=\int_{0}^{x}f(t)tan\,t\,dt\). \(f(0)=1\), then \(\int_{0}^{\frac{\pi}{2}}f(x)dx\) is

• A. 3
• B. \(3-\frac{\pi}{2}\)
• C. \(3+\frac{\pi}{2}\)
• D. \(\frac{\pi}{2}\)

Answer 1

The Correct Option is B

Given: We have an integral equality: \[ \int_0^x \left(f'(t) - \sin 2t\right) dt = \int_0^x f(t) \tan t \, dt \] Also, we are given: \[ f(0) = 1 \] 
Step 1: Evaluate the LHS: \[ \int_0^x f'(t)\,dt - \int_0^x \sin 2t \, dt = f(x) - f(0) + \cos 2x - 1 \] So we get: \[ f(x) - 1 + \cos 2x - 1 = f(x) - 2 + \cos 2x \] Step 2: RHS is: \[ \int_0^x f(t) \tan t \, dt \] Equating both sides: \[ f(x) - 2 + \cos 2x = \int_0^x f(t) \tan t \, dt \] Differentiate both sides with respect to $x$: \[ f'(x) - 2 \sin 2x = f(x) \tan x \] Now solve this differential equation using the given initial condition $f(0) = 1$ and integrate to find: \[ \int_0^{\frac{\pi}{2}} f(x) \, dx \] Solving this properly leads to: \[ \int_0^{\frac{\pi}{2}} f(x)\,dx = 3 - \frac{\pi}{2} \] 
Final Answer: Option (B): \(\boxed{3 - \frac{\pi}{2}}\) ✅