Question

What is the integral of $3\sin \left( 2x \right)$ ?

Answer 1

We need to evaluate $\int{3\sin \left( 2x \right)}dx$. For this, we will use the formulas of integration and proceed step by step to obtain the answer.
The process of finding integrals is known as integration.
Certain rules and formulas of integration will help to solve this problem easily,
$\begin{aligned} & \int{\sin \left( x \right)dx}=-\cos x+c \\\ & \int{A\sin \left( ax+b \right)dx}=A\left( \dfrac{-\cos \left( ax+b \right)}{a} \right)+c \\\ \end{aligned}$

Complete step by step solution:
According to the question, we need to find the value of
$\int{3\sin \left( 2x \right)}dx$
We have the formula,
$\int{A\sin \left( ax+b \right)dx}=A\left( \dfrac{-\cos \left( ax+b \right)}{a} \right)+c$
Comparing it with the given question, we can say that
$\begin{aligned} & A=3 \\\ & a=2 \\\ & b=0 \\\ \end{aligned}$
Since, integral of $\sin x$ is $-\cos x$ , thus using the above expressions, we can write
$\int{3\sin \left( 2x \right)}dx$
Since, 3 is a constant, we can take it out from the integral as it will not affect the integration process.
$=3\int{\sin \left( 2x \right)dx}$
Now, integrating $\sin \left( 2x \right)$ with respect to $x$, we get
$=3\left( \dfrac{-\cos \left( 2x \right)}{2} \right)+c$
Here, we need to check the variable with respect to which we are integrating the expression. Since, we are integrating it with respect to $x$, and here the coefficient of $x$ is 2, so we need to divide the integral by the coefficient of $x$. Thus, we obtain
$=\dfrac{-3\cos \left( 2x \right)}{2}+c$
Hence, the integral of $3\sin \left( 2x \right)$ is given by $\dfrac{-3\cos \left( 2x \right)}{2}+c$, where $c$ is an arbitrary constant value.

Note: While integration, you might have noticed an unknown value c also present in the answer. This is a constant value that arises while indefinite integration. Since the integration is not bound under any limits, we are required to add this unknown value in our answer.