Question

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

Answer 1

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.