Question

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

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

Answer 1

The correct option is D $x+(\pi -2)\sin x$

$f\left(x\right)=x+\underset{0}{\overset{\pi /2}{\int }}\sin x\cdot \cos yf\left(y\right)dy$
Let $\underset{0}{\overset{\pi /2}{\int }}\cos yf\left(y\right)dy=k$
Then $f\left(x\right)=x+k\sin x$
So, $k=\underset{0}{\overset{\pi /2}{\int }}\cos y(y+k\sin y)dy={[y\sin y+\cos y]}_0^{\pi /2}-{\left[\frac{k}{4}\cos 2y\right]}_0^{\pi /2}$
$\Rightarrow k=(\frac{\pi }{2}-1)+\frac{k}{2}$
$\Rightarrow k=\pi -2$
$\text{So}f\left(x\right)=x+(\pi -2)\sin x$