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(B) Differentiating the given equation $\int_0^x (f^{\prime}(t)-\sin 2t) dt = \int_x^0 f(t) 	an t dt$ with respect to $x$ using the Leibniz rule, we

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$3-\frac{\pi}{2}$

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(B) Differentiating the given equation $\int_0^x (f^{\prime}(t)-\sin 2t) dt = \int_x^0 f(t) 	an t dt$ with respect to $x$ using the Leibniz rule, we get:$f^{\prime}(x) - \sin 2x = -f(x) 	an x$$f^{\prime}(x) + f(x) 	an x = \sin 2x$This is a linear differential equation of the form $\frac{df}{dx} + P(x)f = Q(x)$, where $P(x) = 	an x$ and $Q(x) = \sin 2x = 2 \sin x \cos x$.The integrating factor $IF = e^{\int 	an x dx} = e^{\ln |\sec x|} = \sec x$.Multiplying by $IF$, we get $\frac{d}{dx} (f(x) \sec x) = \sin 2x \sec x = 2 \sin x$.Integrating both sides, $f(x) \sec x = \int 2 \sin x dx = -2 \cos x + C$.Given $f(0) = 1$, we have $1 \cdot \sec 0 = -2 \cos 0 + C \implies 1 = -2 + C \implies C = 3$.Thus, $f(x) \sec x = -2 \cos x + 3$, which implies $f(x) = -2 \cos^2 x + 3 \cos x$.Now, $\int_0^{\frac{\pi}{2}} f(x) dx = \int_0^{\frac{\pi}{2}} (-2 \cos^2 x + 3 \cos x) dx$.Using $\cos^2 x = \frac{1+\cos 2x}{2}$, we get $\int_0^{\frac{\pi}{2}} (-1 - \cos 2x + 3 \cos x) dx = [-x - \frac{\sin 2x}{2} + 3 \sin x]_0^{\frac{\pi}{2}}$.$= (-\frac{\pi}{2} - 0 + 3) - (0) = 3 - \frac{\pi}{2}$.

Let $f$ be a non-negative function defined on $\left[0, \frac{\pi}{2}\right]$. If $\int_0^x \left(f^{\prime}(t)-\sin 2t\right) dt = \int_x^0 f(t) \tan t dt$ and $f(0)=1$, then find $\int_0^{\frac{\pi}{2}} f(x) dx$.

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