Integrate: $\dfrac{{\sin x}}{{\sin 2x + 2}}$ with respect to $x$.

Hint:The given question requires us to integrate a function of x with respect to x. We evaluate the given integral using the substitution method. We first simplify the function provided to us in the question and then split it into two parts for the ease of integration. Now, we substitute the denominators of both the rational functions in x as separate variables and do the integration further.

Complete step by step answer:
The given question requires us to integrate the function $\dfrac{{\sin x}}{{\sin 2x + 2}}$.
So, we have, \[\int {\dfrac{{\sin x}}{{\sin 2x + 2}}} dx\]
But, it is very difficult to integrate the function directly. So, we first split the function into two parts after simplifying a bit. So, we get,
\[ \Rightarrow \dfrac{1}{2}\int {\dfrac{{\left( {\sin x + \cos x} \right) + \left( {\sin x - \cos x} \right)}}{{\sin 2x + 2}}} dx\]
\[ \Rightarrow \dfrac{1}{2}\int {\dfrac{{\left( {\sin x + \cos x} \right)}}{{\sin 2x + 2}} + \dfrac{{\left( {\sin x - \cos x} \right)}}{{\sin 2x + 2}}} dx\]

Now, we know that ${\left( {\sin x + \cos x} \right)^2} = \sin 2x + 1$ and ${\left( {\cos x - \sin x} \right)^2} = 1 - \sin 2x$ . So, we get,
\[ \Rightarrow \dfrac{1}{2}\int {\dfrac{{\left( {\sin x + \cos x} \right)}}{{3 - {{\left( {\cos x - \sin x} \right)}^2}}}dx - \dfrac{{\left( {\cos x - \sin x} \right)}}{{1 + {{\left( {\sin x + \cos x} \right)}^2}}}} dx\]
Separating both the integrals, we get,
\[ \Rightarrow \dfrac{1}{2}\int {\dfrac{{\left( {\sin x + \cos x} \right)}}{{3 - {{\left( {\sin x - \cos x} \right)}^2}}}dx - \dfrac{1}{2}\int {\dfrac{{\left( {\cos x - \sin x} \right)}}{{1 + {{\left( {\sin x + \cos x} \right)}^2}}}} } dx\]
Now, we assign $t = \sin x - \cos x$ and $u = \sin x + \cos x$.
So, we have, $dt = \left( {\sin x + \cos x} \right)dx$ and $du = \left( {\cos x - \sin x} \right)dx$
\[ \Rightarrow \dfrac{1}{2}\int {\dfrac{{dt}}{{3 - {t^2}}} - \dfrac{1}{2}\int {\dfrac{{du}}{{1 + {u^2}}}} } \]

Now, we must know the special integral $\int {\dfrac{1}{{{a^2} + {x^2}}}} dx = {\tan ^{ - 1}}\left( {\dfrac{x}{a}} \right) + c$ and \[\int {\dfrac{1}{{{a^2} - {x^2}}} = \dfrac{1}{{2a}}\log \left| {\dfrac{{x + a}}{{x - a}}} \right|} \].
So, using these special integral, we get,
\[ \Rightarrow \dfrac{1}{2}\int {\dfrac{{dt}}{{{{\left( {\sqrt 3 } \right)}^2} - {t^2}}} - \dfrac{1}{2}\int {\dfrac{{du}}{{1 + {u^2}}}} } \]
\[ \Rightarrow \dfrac{1}{2}\left( {\dfrac{1}{{2\sqrt 3 }}\log \left| {\dfrac{{t + \sqrt 3 }}{{t - \sqrt 3 }}} \right|} \right) - \dfrac{1}{2}{\tan ^{ - 1}}u + c\]
Substituting the value of t and u back in the expression, we get,
\[ \Rightarrow \dfrac{1}{2}\left( {\dfrac{1}{{2\sqrt 3 }}\log \left| {\dfrac{{t + \sqrt 3 }}{{t - \sqrt 3 }}} \right|} \right) - \dfrac{1}{2}{\tan ^{ - 1}}u + c\]
\[ \therefore \dfrac{1}{{4\sqrt 3 }}\log \left| {\dfrac{{\sin x - \cos x + \sqrt 3 }}{{\sin x - \cos x - \sqrt 3 }}} \right| - \dfrac{1}{2}{\tan ^{ - 1}}\left( {\sin x + \cos x} \right) + c\]
where $c$ is any arbitrary constant.

So, the integral of $\dfrac{{\sin x}}{{\sin 2x + 2}}$ with respect to x is \[\dfrac{1}{{4\sqrt 3 }}\log \left| {\dfrac{{\sin x - \cos x + \sqrt 3 }}{{\sin x - \cos x - \sqrt 3 }}} \right| - \dfrac{1}{2}{\tan ^{ - 1}}\left( {\sin x + \cos x} \right) + c\] where $c$ is any arbitrary constant.

Note:The indefinite integrals of certain functions may have more than one answer in different forms. However, all these forms are correct and interchangeable into one another. Indefinite integral gives us the family of curves as we don’t know the exact value of the arbitrary constant. We should have an open mind while thinking of a method to solve the given integral.