
How do you find the antiderivative of $\sin (\pi x)dx$ ?
Answer
547.5k+ 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) $\int {co{{\sec }^2}xdx} = - \cot x + C$
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