What is the value of the integral \[\int {\left\{ {\dfrac{{\left( {\sec x \cdot \csc x} \right)}}{{\left( {2\cot x - \sec x \cdot \csc x} \right)}}} \right\}} dx\]?
A. \[\log\left| {\sec x + \tan x} \right| + C\]
B. \[\log\left| {\sec x + \csc x} \right| + C\]
C. \[\dfrac{1}{2}\log\left| {\sec 2x + \tan 2x} \right| + C\]
D. \[\log\left| {\sec 2x + \csc 2x} \right| + C\]
E. \[\log\left| {\sec 2x \cdot \csc 2x} \right| + C\]

Hint First, simplify \[\int {\left\{ {\dfrac{{\left( {\sec x - \csc x} \right)}}{{2\cot x - \sec x - \csc x}}} \right\}} dx\] by putting \[\sec x = \dfrac{1}{{\cos x}}\] , \[\csc x = \dfrac{1}{{\sin x}}\], \[\cot x = \dfrac{{\cos x}}{{\sin x}}\]. Then apply the formula \[2{\cos ^2}x - 1 = \cos 2x\]. After that, substitute \[2x = u\] and derivative it. Then rewrite the integration in terms of \[u\]. Then apply integration formula \[\int {\sec xdx = \log \left| {\sec x + \tan x} \right| + C} \] and substitute \[u = 2x\].

Formula used
Trigonometric ratios:
\[\sec x = \dfrac{1}{{\cos x}}\]
\[\csc x = \dfrac{1}{{\sin x}}\]
\[\cot x = \dfrac{{\cos x}}{{\sin x}}\]
\[\cos 2x = 2\cos^{2} x - 1\]
\[\int {\sec\left( x \right)} dx = \log\left| {\sec x + \tan x} \right| + C\]

Complete step by step solution:
The given integral is \[\int {\left\{ {\dfrac{{\left( {\sec x \cdot \csc x} \right)}}{{\left( {2\cot x - \sec x \cdot \csc x} \right)}}} \right\}} dx\].
Let consider,
\[I = \int {\left\{ {\dfrac{{\left( {\sec x \cdot \csc x} \right)}}{{\left( {2\cot x - \sec x \cdot \csc x} \right)}}} \right\}} dx\]
Let’s simplify the right-hand side of the above equation using the trigonometric ratios.
\[I = \int {\left\{ {\dfrac{{\left( {\dfrac{1}{{\cos x}} \cdot \dfrac{1}{{\sin x}}} \right)}}{{\left( {\dfrac{{2\cos x}}{{\sin x}} - \dfrac{1}{{\cos x}} \cdot \dfrac{1}{{\sin x}}} \right)}}} \right\}} dx\]
Simplify the above integral.
\[I = \int {\left\{ {\dfrac{{\left( {\dfrac{1}{{\sin x \cos x}}} \right)}}{{\left( {\dfrac{{2\cos x}}{{\sin x}} - \dfrac{1}{{\sin x \cos x}}} \right)}}} \right\}} dx\]
Now multiply the numerator and denominator of the first term in denominator by \[\cos x\].
\[I = \int {\left\{ {\dfrac{{\left( {\dfrac{1}{{\sin x \cos x}}} \right)}}{{\left( {\dfrac{{2\cos^{2}x}}{{\sin x \cos x}} - \dfrac{1}{{\sin x \cos x}}} \right)}}} \right\}} dx\]
\[ \Rightarrow \]\[I = \int {\left\{ {\dfrac{{\left( {\dfrac{1}{{\sin x \cos x}}} \right)}}{{\left( {\dfrac{{2\cos^{2}x - 1}}{{\sin x \cos x}}} \right)}}} \right\}} dx\]
Cancel out the common terms.
\[I = \int {\left\{ {\dfrac{1}{{2\cos^{2}x - 1}}} \right\}} dx\]
\[ \Rightarrow \]\[I = \int {\dfrac{1}{{\cos\left( {2x} \right)}}} dx\] [ Since \[\cos 2x = 2\cos^{2}x - 1\]]
\[ \Rightarrow \]\[I = \int {\sec\left( {2x} \right)} dx\] [ Since \[\sec x = \dfrac{1}{{\cos x}}\]]

Apply the substitution method of integration.
Substitute \[2x = u\] in the above integral.
Differentiate \[2x = u\] with respect to \[x\].
\[2 = \dfrac{{du}}{{dx}}\]
\[ \Rightarrow \]\[2dx = du\]

Then,
\[I = \int {\sec\left( u \right)} \dfrac{{du}}{2}\]
Simplify the above integral.
\[I = \dfrac{1}{2}\int {\sec\left( u \right)} du\]
Now apply the standard integral formula \[\int {\sec\left( x \right)} dx = \log\left| {\sec x + \tan x} \right| + C\].
Integrate with respect to \[u\].
\[I = \dfrac{1}{2}\left[ {\log\left| {\sec u + \tan u} \right| + C} \right]\]
\[ \Rightarrow \]\[I = \dfrac{1}{2}\log\left| {\sec u + \tan u} \right| + C\]
Resubstitute the value of \[u\].
\[I = \dfrac{1}{2}log\left| {\sec 2x + \tan 2x} \right| + C\]
Hence the correct option is C.

Note: Students are often forgot to put \[u = 2x\] at the end of the solution. They marked option A is correct. But the correct option is option C.