If $\frac{dy}{dx}+y\tan \mathit{x}=sin2x$ and $y\left(0\right)=1$, then $y\left(\pi \right)$ is equal to

The given differential equation $\frac{dy}{dx}+y\tan x=\sin 2x$ is linear differential equation of the form 

$\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$ with $P\left(x\right)=\tan x$, $Q\left(x\right)=\sin 2x$

Now, $\mathrm{I}.\mathrm{F}.=e^{\int \mathit{P}\left(x\right)dx}=e^{\int \tan x\mathit{dx}}=e^{\ln \left(\sec \mathit{x}\right)}=\sec x$

And, the solution is $y\left(\mathrm{I}.\mathrm{F}.\right)=\int Q\left(x\right)\left(\mathrm{I}.\mathrm{F}.\right)dx+c$

$\Rightarrow y\sec x=\int \sec x\sin 2xdx+c$

$\Rightarrow y\sec x=\int \frac{1}{\cos x}\left(2\sin x\cos x\right)dx+c$

$\Rightarrow y\sec x=\int 2\sin xdx+c$

$\Rightarrow y\sec x=-2\cos x+c$

Given, $y\left(0\right)=1$

$\Rightarrow 1·\sec 0=-2\cos 0+c$

$\Rightarrow 1=-2+c$

$\Rightarrow c=3$

$\Rightarrow y\sec \mathit{x}=-2\cos x+3$

Now, at $x=\pi$

$y\left(\pi \right)\sec \pi =-2\cos \pi +3$

$\Rightarrow y\left(\pi \right)\left(-1\right)=-2\left(-1\right)+3$

$\Rightarrow y\left(\pi \right)=-5.$