Question

How do you evaluate the definite integral $\int {\sin 3xdx} $ from $[0,\pi ]$?

Answer 1

Hint: In this question, we need to find the integration of the given function with respect to x. Here we use a substitution method to evaluate the integration. Firstly, to make integration easier, we take $t = 3x$ and differentiate it. Then we use the obtained expression for $dx$ to integrate the given function. We use the formula of integration of the sine function which is given by, $\int {\sin xdx} = - \cos x + C$ and after that we substitute back $t = 3x$ and simplify to get the desired result.

Complete step by step answer:
Here we are asked to find the antiderivative of the function $\sin (3x)dx$ from $[0,\pi ]$.
i.e. we need to integrate the function $\sin (3x)dx$ from $[0,\pi ]$
So we find out $\int\limits_0^\pi {\sin 3xdx} $ …… (1)
Firstly, we take $3x$ some variable say t and proceed. i.e. take $t = 3x$.
Now differentiating this with respect to x we get,
$ \Rightarrow \dfrac{{dt}}{{dx}} = 3\dfrac{{d(x)}}{{dx}}$
$ \Rightarrow \dfrac{{dt}}{{dx}} = 3$
Now taking $dx$ to the other side, we get,
$ \Rightarrow dt = 3dx$
So the expression for $dx$ is,
$ \Rightarrow dx = \dfrac{{dt}}{3}$
Substituting $t = 3x$ in the equation (1), we get,
$ \Rightarrow \int\limits_0^\pi {\sin 3xdx} = \int\limits_0^\pi {\sin tdx} $
Now put $dx = \dfrac{{dt}}{3}$, we get,
$ \Rightarrow \int\limits_0^\pi {\sin 3xdx} = \int\limits_0^\pi {\sin t\dfrac{{dt}}{3}} $
Since $\dfrac{1}{3}$ is a constant, from the constant coefficient rule we can take it out of integration.
Hence we have,
$ \Rightarrow \int\limits_0^\pi {\sin 3xdx} = \dfrac{1}{3}\int\limits_0^\pi {\sin tdt} $ …… (2)
We know the integration formula of sine function which is given by,
$\int {\sin xdx} = - \cos x + C$, where $C$ is an integration constant.
Since we have definite integral, we do not consider the integration constant.
Hence the equation (2) becomes,
$ \Rightarrow \int\limits_0^\pi {\sin 3xdx} = \dfrac{1}{3}\left[ { - \cos t} \right]_0^\pi $
Substituting back $t = 3x$ we get
$ \Rightarrow \int\limits_0^\pi {\sin 3xdx = \dfrac{1}{3}} \left[ { - \cos 3x} \right]_0^\pi $
Now applying the limits we get,
$ \Rightarrow \int\limits_0^\pi {\sin 3xdx} = - \dfrac{1}{3}\left[ {\cos 3\pi - \cos 3(0)} \right]$
$ \Rightarrow \int\limits_0^\pi {\sin 3xdx} = - \dfrac{1}{3}\left[ {\cos 3\pi - \cos 0} \right]$
$ \Rightarrow \int\limits_0^\pi {\sin 3xdx} = - \dfrac{1}{3}\left[ { - 1 - 1} \right]$
$ \Rightarrow \int\limits_0^\pi {\sin 3xdx} = - \dfrac{1}{3}\left[ { - 2} \right]$
$ \Rightarrow \int\limits_0^\pi {\sin 3xdx} = \dfrac{2}{3}$

Hence, the value of the definite integral of $\int\limits_0^\pi {\sin 3xdx} $ is given by $\dfrac{2}{3}$.

Note: Students must remember that the antiderivative is nothing but integration. And it is important to substitute $3x$ as some variable, since it makes us to integrate easier and also it avoids confusion. While applying the limits we must be careful. First we have to apply the upper limit which is followed by the lower limit.
The integration of some of the trigonometric functions are given below.
(1) $\int {\sin xdx} = - \cos x + C$
(2) $\int {\cos xdx} = \sin x + C$
(3) $\int {{{\sec }^2}xdx} = \tan x + C$