Differential equation $\frac{dy}{dx}-2(3\cos x+5)y=-1$ How can i solve the differential equation 
$$\frac{dy}{dx}-2(3\cos x+5)y=-1$$
What i have try
It represent a linear differential equation of degree and order $1$
So compare with $\frac{dy}{dx}+Py=Q$
We have $P=-2(3\cos x+5)$ and $Q=-1$
And Integrating factor $\text{(I.f)} =e^{\int 2(3\cos x+5)dx}=e^{-2(3\sin x+5x)}$
So solution is $$ y=\int Q\text{(I.f)}dx=-\int e^{-2(3\sin x+5x)}dx$$
How do i solve it Help me please or How can i write its solution . Thanks
Update: wolframalpham alpha show as 
How can i  write  it solution  in that form.
!
 A: You have somehow missed a part of the solution formula. With the integrating factor $\mu(x)=\exp(\int P(x) dx)$ you get
$$
(\mu(x) y(x))'=\mu (x)y'(x)+\mu'(x) y(x)=\mu(x)(y'(x)+P(x)y(x))=\mu(x) Q(x)
\\~\\
\implies
\mu(x)y(x)=\int \mu(x)Q(x)\,dx
$$
which means that in the formula for $y$ you also have to divide by $\mu$. Together with the integration constant that is hidden in the undetermined integral, you get exactly the same solution formula as WA.
A: I started from:
\begin{equation}
e^{\int P(x) d x}\left(\frac{d y}{d x}+P(x) y\right)=Q(x) e^{\int P(x) d x}
\end{equation} Then:
\begin{equation}
\begin{array}{c}
e^{\int P(x) d x}\left(\frac{d y}{d x}+P(x) y\right)=\frac{d}{d x}\left(e^{\int P(x) d x} y\right) \\
\frac{d}{d x}\left(e^{\int P(x) d x} y\right)=Q(x) e^{\int P(x) d x}
\end{array}
\end{equation}
Integrate both sides of the new equation:
\begin{equation}
\int \frac{d}{d x}\left(e^{\int P(x) d x} y\right) d x=\int Q(x) e^{\int P(x) d x} d x
\end{equation}
The Fundamental Theorem of Calculus shows that:
\begin{equation}
\int \frac{d}{d x}\left(e^{\int P(x) d x} y\right) d x=e^{\int P(x) d x} y+C_{1}
\end{equation}
where C1 is an arbitrary constant and RHS also needs to be found and let it
B(x)+ C2, where C2 is a constant due to the integral.
\begin{equation}
e^{\int P(x) d x} y=B(x)+C_{3}
\end{equation}
Divide by the integrating factor to get the solution:
\begin{equation}
y=B(x) e^{-\int P(x) d x}+C_{3} e^{-\int P(x) d x}
\end{equation}
Basically repeating the order will lead you to the solution.
A: Homogeneous Part
$$y'(x)-\underbrace{2(3\cos(x)+5)}_{p(x)}\;y(x)=0$$
Solution: $y(x)=c\exp[P(x)]$ with $P'(x)=p(x)$ and $c\in\mathbb{R}$. In particular $$P(x)=\int p(x)\,\mathrm{d}x=\int [6\cos(x)+10]\, \mathrm{d}x=6\sin{x}+10x$$
Inhomogeneous Part
$$y'(x)-\underbrace{2(3\cos(x)+5)}_{p(x)}y(x)=-1$$
Variation of the constant: $y(x)=c(x)\;\underbrace{\exp[6\sin(x)+10x]}_{h(x)}$. Note that $h(x)$ is a solution of the homogeneous part, thus when you plug into the original equation
$$c'(x)h(x)+c(x)h'(x)-p(x)c(x)h(x)=-1$$
Since $h'(x)=p(x)h(x)$, you get
$$c'(x)=-\frac{1}{h(x)}=-1\exp[-(6\sin(x)+10x)]$$
Integrate:
$$c(x)=c+\int -1\exp[-(6\sin(x)+10x)]\,\mathrm{d}x$$
Put them all together:
\begin{align}
y(x)&=\Big(c+\int -1\exp[-(6\sin(x)+10x)]\,\mathrm{d}x\Big)\exp[6\sin (x)+10x]\\
&=ce^{6\sin(x)+10x}+e^{6\sin(x)+10x}\int e^{-(6\sin(x)+10x)}\,\mathrm{d}x
\end{align}
