Question

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

• A. $x+\frac{2}{3}(\pi -2)\sin x$
• B. $x+(\pi +2)\sin x$
• C. $x+\frac{\pi }{2}\sin x$
• D. $x+(\pi -2)\sin x$

Answer 1

Answer: (D)

$f(x)=x+\underset{0}{\overset{\pi /2}{\int }}\sin x\cos yf(y)dy$
$f(x)=x+\sin x\underset{\text{к }}{\underbrace{\underset{0}{\overset{\pi /2}{\int }}\cos yf(y)dy}}$
$\Rightarrow f(x)=x+K\sin x$
$\Rightarrow f(y)=y+K\sin y$
Now $K=\underset{0}{\overset{\pi /2}{\int }}\cos y(y+K\sin y)dy$
$K=\underset{0}{\overset{\pi /2}{\int }}\underset{\text{Apply IBP}}{\mathrm{ycosdy}}+\underset{0}{\overset{\pi /2}{\int }}\underset{\text{Put sin }yt}{\mathrm{cosysinydy}}$
$K=(y\sin y{)}_0^{\pi /2}-\underset{0}{\overset{\pi /2}{\int }}\sin dy+K{\int }_0^{1}tdt$
$\Rightarrow K=\frac{\pi }{2}-1+K\left(\frac{1}{2}\right)$
$\Rightarrow K=\pi -2$
So $f(x)=x+(\pi -2)\sin x$