Solve the differential equation:
$\dfrac{dy}{dx}+2y\tan x=\sin x$

Hint: Observe that the given differential equation is a linear differential equation, i.e. a differential equation of the form $\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$. The solution of a linear differential equation is given by
$yI.F=\int{Q\left( x \right)I.Fdx}$, where $IF={{e}^{\int{P\left( x \right)dx}}}$. Find the integrating factor of the given differential equation and hence find the solution of the given equation. Verify your answer.

Complete step-by-step answer:
We have $\dfrac{dy}{dx}+2y\tan x=\sin x$, which is of the form of $\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$.
Here $P\left( x \right)=2\tan x$ and $Q\left( x \right)=\sin x$.
Now, we know that the integrating factor IF of a linear differential equation is given by
$IF={{e}^{\int{P\left( x \right)dx}}}$
Hence, the integrating factor of the given differential equation is given by
$IF={{e}^{\int{2\tan xdx}}}$.
We know that $\int{\tan xdx}=\ln \left( \sec x \right)+C$
Hence, we have $IF={{e}^{2\ln \left( \sec x \right)}}={{e}^{\ln \left( {{\sec }^{2}}x \right)}}$
We know that ${{e}^{\ln x}}=x$
Hence, we have
$IF={{\sec }^{2}}x$
Now, we know that the solution of a linear differential equation is given by
$yIF=\int{Q\left( x \right)IFdx}$
Hence, the solution of the given equation is
$y{{\sec }^{2}}x=\int{\sin x{{\sec }^{2}}xdx}$
Now, we have
$\sin x{{\sec }^{2}}x=\sin x\sec x\sec x=\dfrac{\sin x}{\cos x}\sec x=\sec x\tan x$
Hence, we have
$y{{\sec }^{2}}x=\int{\sec x\tan xdx}$
We know that
$\int{\sec x\tan xdx}=\sec x$
Hence, we have
$y{{\sec }^{2}}x=\sec x+C$
Dividing both sides by ${{\sec }^{2}}x$, we get
$y=\cos x+C{{\cos }^{2}}x$, which is the required solution of the given differential equation.

Note: Verification.
We have
$y=\cos x+C{{\cos }^{2}}x$
Differentiating both sides, we get
$\dfrac{dy}{dx}=-\sin x-2C\sin x\cos x$
Adding $2y\tan x$ on both sides, we get
$\dfrac{dy}{dx}+2y\tan x=-\sin x-2C\sin x\cos x+2\tan x\left( \cos x+C{{\cos }^{2}}x \right)$
We have $\tan x\cos x=\dfrac{\sin x}{\cos x}\cos x=\sin x$
Hence, we have
$\dfrac{dy}{dx}+2y\tan x=-\sin x-2C\sin x\cos x+2\sin x+2C\sin x\cos x$
Simplifying, we get
$\dfrac{dy}{dx}+2y\tan x=\sin x$, which is the given differential equation.
Hence our answer is verified to be correct.