Integrate the given function: \[\int{\dfrac{\cos 2x+2{{\sin }^{2}}x}{{{\sin }^{2}}x}}dx\]

Hint: Assume that \[{{I}_{1}}=\int{\dfrac{\cos 2x+2{{\sin }^{2}}x}{{{\sin }^{2}}x}}dx\] . Use the identity, \[\cos 2x=1-2{{\sin }^{2}}x\] and replace \[2{{\sin }^{2}}x\] in the numerator in terms of \[\cos 2x\] . Now, use the identity, \[\sin x=\dfrac{1}{\cos ecx}\] and simplify it. At last, use the formula, \[\int{\cos e{{c}^{2}}xdx}=-\cot x+c\] and solve it further to get the value of \[{{I}_{1}}\].

Complete step-by-step solution:
According to the question, we are asked to find the integration of \[\int{\dfrac{\cos 2x+2{{\sin }^{2}}x}{{{\sin }^{2}}x}}dx\] .
For our convenience, first of all, let us assume that \[{{I}_{1}}=\int{\dfrac{\cos 2x+2{{\sin }^{2}}x}{{{\sin }^{2}}x}}dx\] ……………………………….(1)
We can see that the given expression cannot be integrated directly. So, we have to modify it into a simpler expression that can be integrated directly using the basic formulas.
We know the identity, \[\cos 2x=1-2{{\sin }^{2}}x\] ………………………………………..(2)
On simplifying equation (2) and taking \[\cos 2x\] to the RHS while \[2{{\sin }^{2}}x\] to the LHS, we get
\[\Rightarrow \cos 2x=1-2{{\sin }^{2}}x\]
\[\Rightarrow 2{{\sin }^{2}}x=1-\cos 2x\] ……………………………………(3)
We have to simplify equation (1).
Now, using equation (3) and on substituting \[2{{\sin }^{2}}x\] by \[1-\cos 2x\] in numerator of equation (1), we get
 \[\Rightarrow {{I}_{1}}=\int{\dfrac{\cos 2x+1-\cos 2x}{{{\sin }^{2}}x}}dx\]
\[\Rightarrow {{I}_{1}}=\int{\dfrac{1}{{{\sin }^{2}}x}}dx\] ……………………………………(4)
We also know the identity that sine function is the reciprocal of cosec function, \[\sin x=\dfrac{1}{\cos ecx}\] ………………………………………………(5)
Now, using equation (5) and on substituting \[\sin x\] by \[\cos ecx\] in equation (4), we get
\[\Rightarrow {{I}_{1}}=\int{\dfrac{1}{{{\left( \dfrac{1}{\cos ecx} \right)}^{2}}}}dx\]
\[\Rightarrow {{I}_{1}}=\int{\cos e{{c}^{2}}xdx}\] …………………………………………(6)
We know the formula, \[\int{\cos e{{c}^{2}}xdx}=-\cot x+c\] ………………………………………….(7)
Now, on comparing equation (6) and equation (7), we get
\[\Rightarrow {{I}_{1}}=-\cot x+c\] ………………………………………….(8)
Similarly, on comparing equation (1) and equation (8), we get
\[{{I}_{1}}=\int{\dfrac{\cos 2x+2{{\sin }^{2}}x}{{{\sin }^{2}}x}}dx=-\cot x+c\] …………………………………..(9)
From equation (9), we have got the value of integration, \[\int{\dfrac{\cos 2x+2{{\sin }^{2}}x}{{{\sin }^{2}}x}}dx=-\cot x+c\] .
Therefore, \[\int{\dfrac{\cos 2x+2{{\sin }^{2}}x}{{{\sin }^{2}}x}}dx=-\cot x+c\].

Note: In this question, one might convert the \[{{\sin }^{2}}x\] in the denominator in terms of \[\cos 2x\] by using the identity \[\cos 2x=1-2{{\sin }^{2}}x\] . On doing this, one will get \[\int{\dfrac{2}{1-\cos 2x}}dx\] which is more complex to solve within a given time constraint. So, for simple integration, just convert the \[{{\sin }^{2}}x\] in the numerator in terms of \[\cos 2x\] by using the identity \[\cos 2x=1-2{{\sin }^{2}}x\] and leave the \[{{\sin }^{2}}x\] in the denominator as it is.