How do you integrate \[\dfrac{{\sin x}}{{{{\left( {1 + \sin x} \right)}^2}}}\]?

Hint: To solve this question first we convert the sin function in half-angle in terms of tan function. Then we simplify the function and make a perfect square in the denominator and then put that in another variable then differentiate with respect to x and then put all the values and then split that and integrate that part and then again put that value in the expression and got the right answer we modify that answer if the options are given according to the options.

Complete step by step solution:
We have to integrate \[\dfrac{{\sin x}}{{{{\left( {1 + \sin x} \right)}^2}}}\].
Let, \[I = \int {\dfrac{{\sin x}}{{{{\left( {1 + \sin x} \right)}^2}}}dx} \]
Now we are using the half angle formula in terms of tan function \[\sin x = \dfrac{{2\tan \dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}}\]
Using this formula
\[I = \int {\dfrac{{\dfrac{{2\tan \dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}}}}{{{{\left( {1 + \dfrac{{2\tan \dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}}} \right)}^2}}}dx} \]
Now using the identity of trigonometry \[1 + {\tan ^2}\dfrac{x}{2} = {\sec ^2}\dfrac{x}{2}\]
\[I = \int {\dfrac{{\dfrac{{2\tan \dfrac{x}{2}}}{{{{\sec }^2}\dfrac{x}{2}}}}}{{{{\left( {\dfrac{{{{\sec }^2}\dfrac{x}{2} + 2\tan \dfrac{x}{2}}}{{{{\sec }^2}\dfrac{x}{2}}}} \right)}^2}}}dx} \]
On simplifying the expression.
\[I = \int {\dfrac{{2\tan \dfrac{x}{2}{{\sec }^2}\dfrac{x}{2}}}{{{{\left( {{{\sec }^2}\dfrac{x}{2} + 2\tan \dfrac{x}{2}} \right)}^2}}}dx} \]
Now converting \[{\sec ^2}\dfrac{x}{2}\] in terms of \[{\tan ^2}\dfrac{x}{2}\] using \[1 + {\tan ^2}\dfrac{x}{2} = {\sec ^2}\dfrac{x}{2}\]
\[I = \int {\dfrac{{2\tan \dfrac{x}{2}{{\sec }^2}\dfrac{x}{2}}}{{{{\left( {1 + {{\tan }^2}\dfrac{x}{2} + 2\tan \dfrac{x}{2}} \right)}^2}}}dx} \]
Making a perfect square in the denominator.
\[I = \int {\dfrac{{2\tan \dfrac{x}{2}{{\sec }^2}\dfrac{x}{2}}}{{{{\left( {1 + \tan \dfrac{x}{2}} \right)}^2}}}dx} \]
On putting \[1 + \tan \dfrac{x}{2} = t\] in the equation.
Differentiating the equation both side with respect to x-
\[\dfrac{{{{\sec }^2}\dfrac{x}{2}}}{2}dx = dt\]
Now putting the value in the given equation.
\[I = \int {\dfrac{{4\left( {t - 1} \right)dt}}{{{{\left( t \right)}^2}}}} \]
Now splitting the terms.
\[I = 4\int {\dfrac{{dt}}{t}} - 4\int {\dfrac{1}{{{t^2}}}} dt\]
Now integrating the integrating term.
\[I = 4\ln t + 4\dfrac{1}{t} + c\]
Now putting the value of t again in the equation.
\[I = 4\ln \left( {1 + \tan \dfrac{x}{2}} \right) + \dfrac{4}{{1 + \tan \dfrac{x}{2}}} + c\]
The integration of the expression \[\dfrac{{\sin x}}{{{{\left( {1 + \sin x} \right)}^2}}}\] is-
\[I = 4\ln \left( {1 + \tan \dfrac{x}{2}} \right) + \dfrac{4}{{1 + \tan \dfrac{x}{2}}} + c\]

Note:
This question is hard as well as lengthy also so there are many places where students often make mistakes. To solve this type of question students must have practice and know all the rules and formulas of trigonometry and relations which will help to modify the answer according to the question. We must substitute the value of some expressions and make them simple.