Question

How will you integrate $\int {\sin 2xdx}$ ?

Answer 1

Solve this question by using the substitution method. Substitute $2x$ with any variable and integrate with respect to the variable $x$. Then, after getting the value for $dx$ in terms of $du$ put this in the place of $dx$ in the question.

Complete step by step solution:
We have to integrate the function $\sin 2x$. This can also written as –
$\Rightarrow \int {\sin 2xdx} \cdots \left( 1 \right)$
We normally use the $u$ variable to build new integration in terms of $u$ -
Therefore, now substituting $2x$ for $u$ -
Let $u = 2x$
Differentiating the above equation with respect to $x$, we get –
$\dfrac{{du}}{{dx}} = 2$
By cross – multiplication, we get –
$dx = \dfrac{1}{2}du$
So, now putting $u = 2x$ and $dx = \dfrac{1}{2}du$ in the equation (1), we get –
$\Rightarrow \int {\sin u\dfrac{1}{2}du}$
Now, we can see that, we now have the new integration term which is $u$ but it is the same thing.
In the above integration, we can see that $\dfrac{1}{2}$ is constant value. So, taking $\dfrac{1}{2}$ out of the integration, we get –
$\Rightarrow \dfrac{1}{2}\int {\sin udu}$
We also know that integration is the opposite of differentiation.
So, it is known that the differentiation of $\sin x$ is $\cos x$ therefore, the integration of $\sin x$ is $- \cos x$
Therefore, the integration of $\sin u$ will be –
$\dfrac{1}{2}\int {\sin udu = \dfrac{1}{2}\left( { - \cos u} \right)} \\\ \Rightarrow - \dfrac{1}{2}\left( {\cos u} \right) \\\$
Therefore, now putting the value of $u$ as $2x$ in the above equation, we get –
$\Rightarrow - \dfrac{1}{2}\cos 2x + C$
Hence, after integrating the function $\sin 2x$ or $\int {\sin 2xdx}$ we get the integration as –

$\int {\sin 2xdx} = - \dfrac{1}{2}\cos 2x + C$
So, the above integrated value is the required value.

Note:
We can also integrate this function directly without using the method of substitution. We know that the integration of $\sin x$ is $- \cos x$ and as there is $2x$ in the question so, we have to put the denominator as 2 in the integration of $\sin 2x$. Then, we get the answer as $- \dfrac{{\cos 2x}}{2} + C$.