
How do you find the antiderivative of $\sin (\pi x)dx$ ?
Answer
533.4k+ views
Hint:
In this question, we need to evaluate the antiderivative which is nothing but integration of the given function. Firstly, to make integration easier, we take $t = \pi x$ and differentiate it. Then using the expression of $dx$ obtained, 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 $t = \pi x$ and simplify to get the desired result.
Complete step by step solution:
Here we are asked to find the antiderivative of the function $\sin (\pi x)dx$.
i.e. we need to integrate the function $\sin (\pi x)dx$
So we find out $\int {\sin (\pi x)dx} $ …… (1)
Firstly, we take $\pi x$ some variable say t and proceed. i.e. take $t = \pi x$.
Now differentiating this with respect to x we get,
$ \Rightarrow \dfrac{{dt}}{{dx}} = \pi \dfrac{{d(x)}}{{dx}}$
$ \Rightarrow \dfrac{{dt}}{{dx}} = \pi $
Now taking $dx$ to the other side, we get,
$ \Rightarrow dt = \pi dx$
So the expression for $dx$ is,
$ \Rightarrow dx = \dfrac{{dt}}{\pi }$
Substituting $t = \pi x$ in the equation (1), we get,
$\int {\sin (\pi x)dx} = \int {\sin (t)dx} $
Now put $dx = \dfrac{{dt}}{\pi }$, we get,
$ \Rightarrow \int {\sin (\pi x)dx = \int {\sin t \cdot \dfrac{{dt}}{\pi }} } $
Since $\dfrac{1}{\pi }$ is a constant, from the constant coefficient rule we can take it out of integration.
Hence we have,
$ \Rightarrow \int {\sin (\pi x)dx} = \dfrac{1}{\pi }\int {\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.
Hence the equation (2) becomes,
$ \Rightarrow \int {\sin (\pi x)dx} = \dfrac{1}{\pi }( - \cos t + C)$
$ \Rightarrow \int {\sin (\pi x)dx} = - \dfrac{1}{\pi }\cos t + C$
Substituting back $t = \pi x$ we get,
$ \Rightarrow \int {\sin (\pi x)dx} = - \dfrac{1}{\pi }\cos (\pi x) + C$
Hence, the antiderivative of $\sin (\pi x)dx$ is given by $ - \dfrac{1}{\pi }\cos (\pi x) + C$ , where $C$ is an integration constant.
Note:
Students will get confused about the word antiderivative. But they must remember that, it is nothing but integration. And it is important to substitute $\pi x$ as some variable, since it makes us to integrate easier and also it avoids confusion.
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) $\int {{{\sec }^2}xdx} = \tan x + C$
(4) $\int {co{{\sec }^2}xdx} = - \cot x + C$
In this question, we need to evaluate the antiderivative which is nothing but integration of the given function. Firstly, to make integration easier, we take $t = \pi x$ and differentiate it. Then using the expression of $dx$ obtained, 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 $t = \pi x$ and simplify to get the desired result.
Complete step by step solution:
Here we are asked to find the antiderivative of the function $\sin (\pi x)dx$.
i.e. we need to integrate the function $\sin (\pi x)dx$
So we find out $\int {\sin (\pi x)dx} $ …… (1)
Firstly, we take $\pi x$ some variable say t and proceed. i.e. take $t = \pi x$.
Now differentiating this with respect to x we get,
$ \Rightarrow \dfrac{{dt}}{{dx}} = \pi \dfrac{{d(x)}}{{dx}}$
$ \Rightarrow \dfrac{{dt}}{{dx}} = \pi $
Now taking $dx$ to the other side, we get,
$ \Rightarrow dt = \pi dx$
So the expression for $dx$ is,
$ \Rightarrow dx = \dfrac{{dt}}{\pi }$
Substituting $t = \pi x$ in the equation (1), we get,
$\int {\sin (\pi x)dx} = \int {\sin (t)dx} $
Now put $dx = \dfrac{{dt}}{\pi }$, we get,
$ \Rightarrow \int {\sin (\pi x)dx = \int {\sin t \cdot \dfrac{{dt}}{\pi }} } $
Since $\dfrac{1}{\pi }$ is a constant, from the constant coefficient rule we can take it out of integration.
Hence we have,
$ \Rightarrow \int {\sin (\pi x)dx} = \dfrac{1}{\pi }\int {\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.
Hence the equation (2) becomes,
$ \Rightarrow \int {\sin (\pi x)dx} = \dfrac{1}{\pi }( - \cos t + C)$
$ \Rightarrow \int {\sin (\pi x)dx} = - \dfrac{1}{\pi }\cos t + C$
Substituting back $t = \pi x$ we get,
$ \Rightarrow \int {\sin (\pi x)dx} = - \dfrac{1}{\pi }\cos (\pi x) + C$
Hence, the antiderivative of $\sin (\pi x)dx$ is given by $ - \dfrac{1}{\pi }\cos (\pi x) + C$ , where $C$ is an integration constant.
Note:
Students will get confused about the word antiderivative. But they must remember that, it is nothing but integration. And it is important to substitute $\pi x$ as some variable, since it makes us to integrate easier and also it avoids confusion.
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) $\int {{{\sec }^2}xdx} = \tan x + C$
(4) $\int {co{{\sec }^2}xdx} = - \cot x + C$
Recently Updated Pages
A man running at a speed 5 ms is viewed in the side class 12 physics CBSE

The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

State and explain Hardy Weinbergs Principle class 12 biology CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Which of the following statements is wrong a Amnion class 12 biology CBSE

Differentiate between action potential and resting class 12 biology CBSE

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

Explain sex determination in humans with the help of class 12 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

