How do you integrate $\int{\sqrt{1+\cos 2x}dx}$?

Hint: To solve the given integration, we need to use the cosine double angle trigonometric identity, which is given by $\cos 2\theta =2{{\cos }^{2}}\theta -1$ so that the integral will reduce to $\int{\sqrt{2{{\cos }^{2}}x}}dx$. By using the laws of the radicals and taking out the constants from the integral, we can write the integral as $\sqrt{2}\int{\cos xdx}$ which can be easily solved to get the final result of the integration.

Complete step by step solution:
Let us write the integral given in the above question is as
$\Rightarrow I=\int{\sqrt{1+\cos 2x}dx}$
Now, we know by the cosine double angle formula that $\cos 2\theta =2{{\cos }^{2}}\theta -1$. So that we can substitute $\cos 2x=2{{\cos }^{2}}x-1$ in the above integral to write it as
\[\begin{align}
  & \Rightarrow I=\int{\sqrt{1+\left( 2{{\cos }^{2}}x-1 \right)}dx} \\
 & \Rightarrow I=\int{\sqrt{2{{\cos }^{2}}x}dx} \\
 & \Rightarrow I=\int{\sqrt{2}\sqrt{{{\cos }^{2}}x}dx} \\
\end{align}\]
Taking the constant \[\sqrt{2}\] outside of the above integral, we get
$\Rightarrow I=\sqrt{2}\int{\sqrt{{{\cos }^{2}}x}dx}$
On cancelling the square with the square root, we get
$\Rightarrow I=\sqrt{2}\int{\cos xdx}$
Now, we know that the integration of the cosine function is equal to the sine function. Therefore, the above integral becomes
\[\Rightarrow I=\sqrt{2}\sin x+C\]

Hence, we have finally evaluated the integral given in the above question as \[\sqrt{2}\sin x+C\].

Note: Since the integration given in the above question is indefinite, do not forget to put a constant at the end. For solving these types of integrations, we must be familiar with the different trigonometric identities. Also, we must remember the integrations of various trigonometric functions for solving these types of questions. We must not be confused between the derivative and the integration of the trigonometric functions. The derivative of cosine is equal to the negative of sine, but the integration is equal to sine.