The function f(x),that satisfies the conditio | vedclass

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(D) Given the integral equation: $f(x)=x+\int_{0}^{\pi / 2} \sin x \cos y f(y) dy$. Since $\sin x$ is independent of $y$,we can write: $f(x)=x+\sin x

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$x+(\pi-2) \sin x$

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(D) Given the integral equation: $f(x)=x+\int_{0}^{\pi / 2} \sin x \cos y f(y) dy$. Since $\sin x$ is independent of $y$,we can write: $f(x)=x+\sin x \int_{0}^{\pi / 2} \cos y f(y) dy$. Let $K = \int_{0}^{\pi / 2} \cos y f(y) dy$. Then $f(x) = x + K \sin x$. Substituting $f(y) = y + K \sin y$ into the expression for $K$: $K = \int_{0}^{\pi / 2} \cos y (y + K \sin y) dy = \int_{0}^{\pi / 2} y \cos y dy + K \int_{0}^{\pi / 2} \sin y \cos y dy$. Using integration by parts for the first integral: $\int y \cos y dy = y \sin y - \int \sin y dy = y \sin y + \cos y$. Evaluating from $0$ to $\pi/2$: $[y \sin y + \cos y]_{0}^{\pi/2} = (\frac{\pi}{2} \cdot 1 + 0) - (0 + 1) = \frac{\pi}{2} - 1$. For the second integral: $\int_{0}^{\pi / 2} \sin y \cos y dy = [\frac{\sin^2 y}{2}]_{0}^{\pi/2} = \frac{1}{2}$. Thus,$K = (\frac{\pi}{2} - 1) + K(\frac{1}{2})$. $K - \frac{K}{2} = \frac{\pi}{2} - 1 \Rightarrow \frac{K}{2} = \frac{\pi-2}{2} \Rightarrow K = \pi - 2$. Substituting $K$ back into $f(x)$: $f(x) = x + (\pi - 2) \sin x$.

The function $f(x)$,that satisfies the condition $f(x)=x+\int_{0}^{\pi / 2} \sin x \cdot \cos y f(y) dy$,is :

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