Question

How do you find the integral of ${\sin ^2}2x$ ?

Answer 1

Hint:
Use the cosine double angle formula to simplify ${\sin ^2}2x$ . Double the angle of $\cos $ in cosine double angle formula and we will get the required formula to convert $\sin 2x$ in terms of $\cos $. By using this identity, we can also bypass the power of 2 in ${\sin ^2}2x$.

Complete step by step solution:
From the question, we know that, we have to find the integration of ${\sin ^2}2x$ which can be expressed mathematically as –
$\int {{{\sin }^2}2xdx} \cdots \left( 1 \right)$
Now, we know that, cosine double angle formula is –
$\cos 2x = 2{\cos ^2}x - 1$, and $\cos 2x = 1 - 2{\sin ^2}x$
So, we have to choose which identity we should choose from the above identities.
Therefore, the identity $\cos 2x = 2{\cos ^2}x - 1$ have only $\cos $ terms so, it cannot be used in this question as there is $\sin $ present in this question and we have to convert it in the terms of $\cos $. So, we will use the identity $\cos 2x = 1 - 2{\sin ^2}x$ as it has both $\sin $ and $\cos $.
In the question, we have been given with ${\sin ^2}2x$, therefore, in the identity $\cos 2x = 1 - 2{\sin ^2}x$ double the angle so, we can get the term ${\sin ^2}2x$ in that identity –
$\cos 4x = 1 - 2{\sin ^2}2x$
Hence, now to convert ${\sin ^2}2x$ in the form of $\cos x$ we have to use the transposition to use the above identity –
$ \Rightarrow {\sin ^2}2x = \dfrac{{1 - \cos 4x}}{2}$
Therefore, now, putting the above value of ${\sin ^2}2x$ in the equation (1), we get –
$ \Rightarrow \int {\dfrac{{1 - \cos 4x}}{2}dx} $
Taking constant $\dfrac{1}{2}$ out of the integration, we get –
\[ \Rightarrow \dfrac{1}{2}\int {\left( {1 - \cos 4x} \right)} dx\]
Now, separating the integration for $1$ and $\cos 4x$, we get –
$ \Rightarrow \dfrac{1}{2}\left[ {\int {1dx - \int {\cos 4xdx} } } \right]$
We know that, $\int {1dx = x} $ and $\int {\cos xdx = \sin x} $
Hence, now integrating with respect to $x$ , we get –
$ \Rightarrow \dfrac{1}{2}\left[ {x - \dfrac{{\sin 4x}}{4}} \right] + C$
By further solving, we get –
$ \Rightarrow \dfrac{x}{2} - \dfrac{{\sin 4x}}{8} + C$

Hence, the integration of ${\sin ^2}2x$ is $\dfrac{x}{2} - \dfrac{{\sin 4x}}{8} + C$.

Note:
Many students go wrong while using the suitable identity for ${\sin ^2}2x$, many of them use the identity $\sin 2x = 2\sin x\cos x$ which will give the wrong answer and be hard to solve. The trigonometric identities should be remembered by the students, so that they can use them suitably according to the question.