If f^{prime}(x)=a cos x+b sin x and f^{prime} | vedclass

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(B) Given $f^{\prime}(x)=a \cos x+b \sin x$ ... $(i)$ Integrating both sides with respect to $x$: $f(x) = \int (a \cos x + b \sin x) dx = a \sin x - b

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$4 \cos x+2 \sin x+1$

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(B) Given $f^{\prime}(x)=a \cos x+b \sin x$ ... $(i)$ Integrating both sides with respect to $x$: $f(x) = \int (a \cos x + b \sin x) dx = a \sin x - b \cos x + C$ ... $(ii)$ Given $f^{\prime}(0) = 4$: $f^{\prime}(0) = a \cos(0) + b \sin(0) = a(1) + b(0) = a = 4$. Given $f(0) = 3$: $f(0) = a \sin(0) - b \cos(0) + C = 0 - b(1) + C = -b + C = 3$. Given $f\left(\frac{\pi}{2}\right) = 5$: $f\left(\frac{\pi}{2}\right) = a \sin\left(\frac{\pi}{2}\right) - b \cos\left(\frac{\pi}{2}\right) + C = a(1) - b(0) + C = a + C = 5$. Since $a = 4$,we have $4 + C = 5 \Rightarrow C = 1$. Substituting $C = 1$ into $-b + C = 3$: $-b + 1 = 3 \Rightarrow -b = 2 \Rightarrow b = -2$. Substituting $a = 4, b = -2, C = 1$ into equation $(ii)$: $f(x) = 4 \sin x - (-2) \cos x + 1 = 4 \sin x + 2 \cos x + 1$.

If $f^{\prime}(x)=a \cos x+b \sin x$ and $f^{\prime}(0)=4, f(0)=3, f\left(\frac{\pi}{2}\right)=5$,then $f(x)=$

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