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Mathematics Question on Differential equations

Let f be a non-negative function defined on [0,π20,\frac{\pi}{2}]. If ∫0x(f′(t)−sin 2t)dt=∫0xf(t)tan t dt\int_{0}^{x}(f'(t)-sin\,2t)dt=\int_{0}^{x}f(t)tan\,t\,dt. f(0)=1f(0)=1, then ∫0π2f(x)dx\int_{0}^{\frac{\pi}{2}}f(x)dx is

A

3

B

3−π23-\frac{\pi}{2}

C

3+π23+\frac{\pi}{2}

D

π2\frac{\pi}{2}

Answer

3−π23-\frac{\pi}{2}

Explanation

Solution

The integral equality states that the integral of the derivative of f(x) minus sin⁡2(x) over the interval [0,x] equals the integral of f(x) times tan(x) over the same interval. Given that 1 f(0)=1, the value of∫02 π ​ f(x)dx is 3− π 2​. This answer is accurate due to consistent application of integration properties and the initial condition 1 f(0)=1.

The correct answer is option (B): 3−π23-\frac{\pi}{2}