Question

How to use u substitution for $\sin 2x$?

Answer 1

Hint: In this question they have given $\sin 2x$ and asked us to solve it by using u substitution. We will take the variable as u and find the derivative of it and rearrange it to substitute in the original integral to make the answer come out easier. After integrating it, we have to replace the u with the real value.

Formulas used: $\int {\sin x = - \cos x} $
$\int {\sin 2x = \dfrac{{ - \cos 2x}}{2}} $

Complete step by step answer:
We are given with a trigonometric function and asked to find the integral of it by using the u substitution method.
Given I =\[\smallint \sin 2x\;dx\]
Since we cannot take the trigonometric function as a whole as u because there is just one trigonometric function is given, we will take the value of the trigonometric function given in variables.
Let the function $2x$ be $u$,
Derivation of $u$with respect to $x$ is done as follow,$u = 2x$
$ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{{d\left( {2x} \right)}}{{dx}}$
$ \Rightarrow \dfrac{{du}}{{dx}} = 2$
Rearranging it we get,
\[ \Rightarrow du = 2\;dx\]
Finding the value of $dx$ ,
 \[ \Rightarrow dx = \dfrac{{du}}{2}\]
Now, as we know the value of $u$ and $dx$
That is $u = 2x$ and \[dx = \dfrac{{du}}{2}\]
We will substitute it in the original integral, then we get,
\[ \Rightarrow \smallint \sin 2x\;dx\]
\[I = \smallint \sin u\;\dfrac{{du}}{2}\]
Taking the numbers parts i.e., $\dfrac{1}{2}$ outside the integral,
\[ \Rightarrow \dfrac{1}{2}\smallint \sin u\;du\]
As we know $\int {\sin x = - \cos x} $ we get,
\[ \Rightarrow \dfrac{1}{2} \times - \cos u + C\]
\[ \Rightarrow - \dfrac{1}{2}\cos u + C\]
Replacing back the value of $u$ that is \[u = 2x\] , we get:
\[ \Rightarrow - \dfrac{1}{2}\cos 2x + C\]
Therefore \[ - \dfrac{1}{2}\cos 2x + C\] is the required answer.

Note: Alternative method:
There is another method to find \[\smallint \sin 2x\;dx\] without using the substitution method.
If question is not clear about the method to be used, you can use this: \[\smallint \sin 2x\;dx\]
Since we know that,$\int {\sin 2x = \dfrac{{ - \cos 2x}}{2}} $,
We will first integrate the 1st function and then the 2nd function which is the variable $2x$.
Therefore it becomes, \[\smallint \sin 2x\;dx\]
\[ \Rightarrow \dfrac{{ - \cos 2x}}{2}dx\]