What is the value of integral $\int{\left( 1+\cos 2x \right)dx}$?

Hint: Assume the given integral as I. Now, break the integral into two parts separated by the plus (+) sign. For the first part of the integral write the constant 1 as ${{x}^{0}}$ and use the formula $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$ to find its anti – derivative. Now, for the second part of the integral apply the formula for the integral of the cosine function given as \[\int{\cos \left( ax+b \right)dx}=\dfrac{\sin \left( ax+b \right)}{a}\]. Add the constant of indefinite integral (c) in the end to complete the answer.

Complete step-by-step solution:
Here we are asked to find the integral of the function \[1+\cos 2x\]. Let us assume the integral as I, so we have,
$I=\int{\left( 1+\cos 2x \right)dx}$
Breaking the integral into two parts we get,
\[\Rightarrow I=\int{1dx}+\int{\cos 2xdx}\]
Here we can see that in the first part of the integral we have a constant function while in the second part we have a trigonometric function, so let us find the integral of these parts one by one. Now, we can write 1 as ${{x}^{0}}$, so we get,
\[\Rightarrow I=\int{{{x}^{0}}dx}+\int{\cos 2xdx}\]
Using the formula $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$ we get,
\[\begin{align}
  & \Rightarrow I=\dfrac{{{x}^{0+1}}}{0+1}+\int{\cos 2xdx} \\
 & \Rightarrow I=\dfrac{x}{1}+\int{\cos 2xdx} \\
 & \Rightarrow I=x+\int{\cos 2xdx} \\
\end{align}\]
Now, using the formula for the anti – derivative of a cosine function having a linear argument given as \[\int{\cos \left( ax+b \right)dx}=\dfrac{\sin \left( ax+b \right)}{a}\] we get,
\[\therefore I=x+\dfrac{\sin \left( 2x \right)}{2}+c\]
Here ‘c’ is the constant of integration as we are evaluating an indefinite integral. Hence, the above relation is our answer.

Note: Note that here you must not use the half angle formula of the cosine function given as $\left( 1+\cos 2x \right)=2{{\cos }^{2}}x$ to simplify the function inside integral because we do not have a direct formula for the integral of the function ${{\cos }^{2}}x$. Also, remember that you cannot apply the formula $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$ for n = -1 because in that case we have the function $\dfrac{1}{x}$ whose integral is $\ln x$. Remember the formulas of integral and differential of all the trigonometric functions.