Question

What is the integral of $\int{{{\sin }^{2}}\left( 2x \right)dx}$ ?

Answer 1

Hint: We need to find the integral of the function ${{\sin }^{2}}\left( 2x \right)$ . We start to solve the question by simplifying the given function in terms of cosine function using the trigonometric formulae. Then, we evaluate the integral of the simplified function using the integration formulae to get the desired result.

Complete step by step solution:
Let $I$ be the value of the integral for the given function.
$\Rightarrow I=\int{{{\sin }^{2}}\left( 2x \right)}dx$
We are given a function and need to integrate it. We solve this question using the trigonometric formulae to simplify the function and then find the value of $I$ .
According to the question,
The integral of the function ${{\sin }^{2}}\left( 2x \right)$ is written as follows,
$\Rightarrow I=\int{{{\sin }^{2}}\left( 2x \right)}dx$
We need to simplify the given trigonometric function.
From trigonometry,
We know that $\cos \left( 2x \right)={{\cos }^{2}}x-{{\sin }^{2}}x$
Applying the same for the given function, we get,
$\Rightarrow \cos \left( 4x \right)={{\cos }^{2}}\left( 2x \right)-{{\sin }^{2}}\left( 2x \right)$
We need to write ${{\cos }^{2}}\left( 2x \right)$ in terms of ${{\sin }^{2}}\left( 2x \right)$
From trigonometric identities,
We know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$
Similarly,
$\Rightarrow {{\sin }^{2}}\left( 2x \right)+{{\cos }^{2}}\left( 2x \right)=1$
Shifting ${{\sin }^{2}}\left( 2x \right)$ to another side of the equation, we get,
$\Rightarrow {{\cos }^{2}}\left( 2x \right)=1-{{\sin }^{2}}\left( 2x \right)$
Substituting the value of ${{\cos }^{2}}\left( 2x \right)$ in $\cos \left( 4x \right)={{\cos }^{2}}\left( 2x \right)-{{\sin }^{2}}\left( 2x \right)$ , we get,
$\Rightarrow \cos \left( 4x \right)=\left( 1-{{\sin }^{2}}\left( 2x \right) \right)-{{\sin }^{2}}\left( 2x \right)$
Simplifying the above equation, we get,
$\Rightarrow \cos \left( 4x \right)=1-{{\sin }^{2}}\left( 2x \right)-{{\sin }^{2}}\left( 2x \right)$
$\Rightarrow \cos \left( 4x \right)=1-2{{\sin }^{2}}\left( 2x \right)$
Finding the value of ${{\sin }^{2}}\left( 2x \right)$ from the above equation, we get,
$\Rightarrow {{\sin }^{2}}\left( 2x \right)=\dfrac{\left( 1-\cos \left( 4x \right) \right)}{2}$
Substituting the value of ${{\sin }^{2}}\left( 2x \right)$ in the integral,
$\Rightarrow I=\int{\dfrac{\left( 1-\cos 4x \right)}{2}}dx$
Taking the value of $\dfrac{1}{2}$ out of the integral, we get,
$\Rightarrow I=\dfrac{1}{2}\int{\left( 1-\cos 4x \right)}dx$
Splitting the integral between the terms, we get,
$\Rightarrow I=\dfrac{1}{2}\int{1}dx-\dfrac{1}{2}\int{\cos 4x}dx$
From the formulae of integration,
$\int{1dx=x}+c$
$\int{\cos ax=\dfrac{\sin ax}{a}}+c$
Similarly,
$\int{\cos 4x=\dfrac{\sin 4x}{4}}+c$
Substituting the same, we get,
$\Rightarrow I=\left( \dfrac{1}{2}\times x \right)-\dfrac{1}{2}\left( \dfrac{\sin 4x}{4} \right)$
Simplifying the above equation, we get,
$\Rightarrow I=\dfrac{x}{2}-\dfrac{\sin 4x}{8}+c$
Substituting the value of $I$ in the above equation,
$\therefore \int{{{\sin }^{2}}\left( 2x \right)dx}=\dfrac{x}{2}-\dfrac{\sin 4x}{8}+c$

Note: We must know the trigonometric identities and formulae to solve the given question easily. One must always remember that the value of the integral $\cos ax$ is $\dfrac{\sin ax}{a}$ and not $\sin ax$. We should always write the functions ${{\sin }^{2}}\left( x \right),{{\cos }^{2}}\left( x \right)$ in terms of cosine function for easy evaluation of integral.