Solve \[\int {\dfrac{{\sin (2x)}}{{2{{\cos }^2}(x) + {{\sin }^2}(x)}}dx} = \]
A. \[ - \ln |{\sin ^2}(x) + 1| + c\]
B. \[\ln |{\cos ^2}(x) + 1| + c\]
C. \[ - \ln |{\cos ^2}(x) + 1| + c\]
D. \[\ln |1 + {\tan ^2}(x)| + c\]

Hint: This question involves multiple concepts like trigonometry and integration by substitution. These types are questions are easily solved by substituting the values to some functions, which simplify the whole equation.

Formula used:
The formulas involved in this question are:
\[{\cos ^2}(\theta ) + {\sin ^2}(\theta ) = 1\], where \[\theta \] can be any real number.
\[\Rightarrow \sin (2\theta ) = 2\sin (\theta )\cos (\theta )\], where \[\theta \] can be any real number.
\[\Rightarrow \int {\dfrac{{dx}}{x} = \ln |x|} + c\], where $c$ is an arbitrary constant.


Complete step by step answer:
 Let us begin this question with the function we are given to integrate, i.e.,
\[ \Rightarrow \int {\dfrac{{\sin (2x)}}{{2{{\cos }^2}(x) + {{\sin }^2}(x)}}dx} \]
Now, by applying the trigonometric identity, \[{\cos ^2}(\theta ) + {\sin ^2}(\theta ) = 1\] we get,
\[ \Rightarrow \int {\dfrac{{\sin (2x)}}{{{{\cos }^2}(x) + 1}}dx} \]
Now, applying the trigonometric identity, \[\sin (2\theta ) = 2\sin (\theta )\cos (\theta )\] we get,
\[ \Rightarrow \int {\dfrac{{2\sin (x)\cos (x)}}{{{{\cos }^2}(x) + 1}}dx} \]
Now, let us make a substitution as shown below,
\[ \Rightarrow t = \cos (x)\]
Now, by differentiating both sides of the equation we get,
\[ \Rightarrow dt = - \sin (x)dx\]

Now, by replacing the values of cos(x) and dx in the integral we get,
\[ \Rightarrow \int {\dfrac{{2\sin (x)\cos (x)}}{{{{\cos }^2}(x) + 1}}dx} = - \int {\dfrac{{2t}}{{{t^2} + 1}}dt} \]
Again, making a substitution as shown below,
\[ \Rightarrow z = {t^2} + 1\]
Now, by differentiating both sides of the equation we get,
\[ \Rightarrow dz = 2tdt\]
Now, by replacing the value of z = t2+1 and dz = 2tdt we get,
\[ \Rightarrow - \int {\dfrac{{2t}}{{{t^2} + 1}}dt} = - \int {\dfrac{{dz}}{z}} \]

Now, the question is converted into an elementary form and can be differentiated using a primary method as shown below,
\[ \Rightarrow - \int {\dfrac{{dz}}{z}} = - \ln |z| + c\]
Now, by replacing the value of z in terms of t we get,
\[ \Rightarrow - \ln |{t^2} + 1| + c\]
And, by finally, replacing the value of t in terms of cos(x) we get,
\[ \therefore - \ln |{\cos ^2}(x) + 1| + c\]Where c is an arbitrary constant.

Thus, option C is the correct answer.

Note:The integration denotes the summation of discrete data. The integral is calculated to find the functions which will describe the area, displacement, volume, that occurs due to a collection of small data, which cannot be measured singularly. In a broad sense, in calculus, the idea of limit is used where algebra and geometry are implemented.