Question

A function $f\left(x\right)$ satisfies $f\left(x\right)=\sin x+{\int }_0^x{f}^{\prime }\left(t\right)(2\sin t-{\sin }^{2}t)dt$, then $f\left(x\right)$ is

• A. $\frac{x}{1-\sin x}$
• B. $\frac{\sin x}{1-\sin x}$
• C. $\frac{1-\cos x}{\cos x}$
• D. $\frac{\tan x}{1-\sin x}$

Answer 1

Answer: (B)

$\frac{\sin x}{1-\sin x}$

$f\left(x\right)=\sin x+{\int }_0^x{f}^{\prime }\left(t\right)(2\sin t-{\sin }^{2}t)dt$
put $x=0$, we get $f\left(0\right)=0$
now differentiating $f\left(x\right)$, we get
$\Rightarrow f^{\prime }\left(x\right)=\cos x+f^{\prime }\left(x\right)(2\sin x-{\sin }^{2}x)$
$\Rightarrow f^{\prime }\left(x\right)=\frac{\cos x}{{\sin }^{2}x-2\sin x+1}=\frac{\cos x}{{(\sin x-1)}^2}$
by integrating $f\left(x\right)=-\frac{1}{\sin x-1}+C$
put $x=0$
$f\left(0\right)=-\frac{1}{0-1}+C=0$
$\Rightarrow C=-1$
Therefore, $f\left(x\right)=-\frac{1}{\sin x-1}-1=\frac{\sin x}{\sin x-1}$
Ans: B