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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 :

Similar Questions

Integrate the function: $\frac{x+2}{\sqrt{x^{2}+2 x+3}}$

If $I = \int \frac{dx}{\sin(x-a) \sin(x-b)}$,then $I$ is given by

If $I_n = \int_{\pi / 2}^{\infty} e^{-x} \cos^n x \, dx$,then $\frac{I_{2018}}{I_{2016}} = $

$\int \frac{dx}{2+\cos x} = $ (Where $C$ is a constant of integration.)

$\text{If } \int x[\log (1+x)]^3 dx = \frac{(1+x)^2}{16}(f(x)) + (1+x)(g(x)), \text{ then } f(x) + g(x) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module