Question

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

$f(x)=x+\int\limits_{0}^{\pi / 2} \sin x \cos y f(y) d y$
$f(x)=x+\sin x \underbrace{\int\limits_{0}^{\pi / 2} \cos y f(y) d y}_{\text {к }}$
$\Rightarrow f(x)=x+K \sin x$
$\Rightarrow f(y)=y+K \sin y$
Now $K=\int\limits_{0}^{\pi / 2} \cos y(y+K \sin y) d y$
$K=\int\limits_{0}^{\pi / 2} \underset{\text{Apply IBP}}{y \cos d y}+\int\limits_{0}^{\pi / 2} \underset{\text{Put sin }y t}{\cos y \sin y d y}$
$K=(y \sin y)_{0}^{\pi / 2}-\int\limits_{0}^{\pi / 2} \sin d y+K \int_{0}^{1} t d t$
$\Rightarrow K =\frac{\pi}{2}-1+ K \left(\frac{1}{2}\right)$
$\Rightarrow K =\pi-2$
So $f(x)=x+(\pi-2) \sin x$