Question

Find all functions f(x) defined on $(-\frac{\pi }{2},\frac{\pi }{2})$ with real values and has a primitive F(x) such that $f\left(x\right)+cosx.F\left(x\right)=\frac{sin2x}{(1+sinx{)}^2}$. Then find f(x).

• A. $f\left(x\right)=-\frac{2cosx}{(1+sinx{)}^2}-C{e}^{-cosx}.sinx$
• B. $f\left(x\right)=-\frac{2cosx}{(1+sinx{)}^2}-C{e}^{-sinx}.cosx$
• C. $f\left(x\right)=-\frac{2cosx}{(1+cosx{)}^2}-C{e}^{sinx}.sinx$
• D. $f\left(x\right)=-\frac{2cosx}{(1+sinx{)}^2}-C{e}^{cosx}.cosx$

Answer 1

The correct option is B $f\left(x\right)=-\frac{2cosx}{(1+sinx{)}^2}-C{e}^{-sinx}.cosx$

$\int f\left(x\right)=F\left(x\right)+c\Rightarrow f\left(x\right)=F^{\prime }\left(x\right)$

Let $F\left(x\right)=y\Rightarrow F^{\prime }\left(x\right)=\frac{dy}{dx}$

$\therefore f\left(x\right)+\cos x.F\left(x\right)=\frac{\sin 2x}{{(1+\sin x)}^2}$

$\Rightarrow \frac{dy}{dx}+\cos x.y=\frac{\sin 2x}{{(1+\sin x)}^2}$ ....(1)

Here $P=\cos x\Rightarrow \int PdP=\int \cos xdx=\sin x$

$\therefore I.F.=e^{\sin x}$

Multiplying (1) by $I.F.$, we get

$e^{\sin x}.\frac{dy}{dx}+e^{\sin x}.\cos x.y=e^{\sin x}.\frac{\sin 2x}{{(1+\sin x)}^2}$

Integrating both sides, we get

$y.e^{\sin x}=2\int e^{\sin x}.\frac{\sin 2x}{{(1+\sin x)}^2}+c$

Put $\sin x=t\Rightarrow \cos xdx=dt$

$\Rightarrow y.e^t=\int \frac{e^t(t+1-1)}{{(1+t)}^2}dt=2\int e^t\{\frac{1}{t+1}+\frac{-1}{{(1+t)}^2}\}+c$

$\Rightarrow y{e}^t=\frac{2e^t}{t+1}+c\Rightarrow y{e}^{sinx}=\frac{2e^{sinx}}{sinx+1}+c\Rightarrow y=\frac{2}{sinx1}+c.e^{-sinx}$

$\therefore f\left(x\right)=\frac{dy}{dx}=\frac{-2\cos x}{{(\sin x+1)}^2}-C.e^{-\sin x}.\cos x$