Question

If $f\left(x\right)=\sin x+{\int }_0^x{f}^{\prime }\left(t\right)(2\sin t-{\sin }^{2}t)dt$ 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. $\sin x$

Answer 1

The correct option is B $\frac{\sin x}{1-\sin x}$

$f^{\prime }\left(x\right)=\cos x+f^{\prime }\left(x\right)(2\sin x-{\sin }^{2}x)$
$\frac{dy}{dx}=\frac{\cos x}{(1-\sin x{)}^2}\Rightarrow \int dy=\int \frac{\cos x}{(1-\sin x{)}^2}dx$
$y=-\frac{1}{(\sin x-1)}+c$
$f\left(0\right)=0\therefore 0=1+c\Rightarrow c=-1$
$y=-1+\frac{1}{1-\sin x}$
$\Rightarrow y=\frac{\sin x}{1-\sin x}$