Solve the given trigonometric functions $\int {{{\sin }^2}\left( {2x + 1} \right)dx} $

Hint: In the above integral question we have to solve the integral so here we will use the simple trigonometric formula for solving it so that we can apply the formula, we have to make some changes in the equation without affecting the values. Also, we need some trigonometric relations to solve them. Then, with the use of basic integration, we can reach the answer very easily.

Formula used:
${\sin ^2}\theta = \dfrac{{1 - \cos 2\theta }}{2}$
Here,
$\theta $, will be the angle made by them.

Complete step-by-step answer:
We have the integral
$I = \int {{{\sin }^2}\left( {2x + 1} \right)dx} $
So it can also be written as
$ \Rightarrow I = \int {\dfrac{1}{2}2{{\sin }^2}\left( {2x + 1} \right)dx} $
Now, we will take the constant term outside and applying the formula, we get
$ \Rightarrow I = \dfrac{1}{2}\int {1 - \cos 2\left( {2x + 1} \right)dx} $
So we will not extract the integrals to integrate it independently, we get
$ \Rightarrow I = \dfrac{1}{2}\int {dx - \dfrac{1}{2}\int {\cos \left( {4x + 2} \right)} dx} $
Now on integrating the equation, we get
And as we know the integration of constant is $x$here, so it can be written as
$ \Rightarrow I = \dfrac{x}{2} - \dfrac{1}{8}\int {\cos \left( {4x + 2} \right)} dx$
Again integrating the above, we get
$ \Rightarrow I = \dfrac{x}{2} - \dfrac{{\sin \left( {4x + 2} \right)}}{8} + c$
And taking the LCM, we get on solving
$ \Rightarrow I = \dfrac{{4x - \sin \left( {4x + 2} \right)}}{8} + c$
Thus, this is the necessary value of the integral.
Hence, this is the necessary answer.

Additional information:
Integration is a method of adding cuts to locate the entirety. Joining can be utilized to discover regions, volumes, essential issues, and numerous valuable things. Yet, it is least demanding to begin with finding the area under the bend of a capacity.

Note: At whatever point we face such kinds of issues the key idea included is to improve within the substance of the integration to the essential level so the immediate integration equation for geometrical terms could be applied straightforwardly. This will assist you with jumping and progressing nicely to arrive at the appropriate response.