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Question: What is \(\int{{{\left( \sin (2\pi x) \right)}^{2}}dx}\) ?...

What is ∫(sin⁡(2πx))2dx\int{{{\left( \sin (2\pi x) \right)}^{2}}dx} ?

Explanation

Solution

‘Integral is also known as the antiderivative’
An antiderivative, inverse derivative, primitive function, primitive integral or indefinite integral of a function f is a differentiable function F whose derivative is equal to the original function f. This can be stated symbolically as F' = f.

Complete step by step solution:
The process of solving for antiderivatives is called antidifferentiation (or indefinite integration), and its opposite operation is called differentiation, which is the process of finding a derivative. Antiderivatives are often denoted by capital Roman letters such as F and G.
Antiderivative can also be used to compute the definite integrals by using the fundamental theorems of calculus. Antiderivative is commonly asked as to evaluate the indefinite integral of any function.
We see that antiderivatives of elementary function are considerably harder than just evaluating their derivatives. We know and can explore many properties of antiderivatives in order to solve quickly and get handy with these. The linearity of integration basically removes the complex integrand into simpler ones and hence it becomes easy for us to evaluate majorly in the case of definite integrals where we are provided with the limits that we need to substitute at the place of the dependent variable and hence get the definite answer.
Now in the question we need to find the integral of I=∫(sin⁡(2πx))2dxI=\int{{{\left( \sin (2\pi x) \right)}^{2}}dx}
So, for this let us use the identity that sin⁡2x=1−cos⁡2x2{{\sin }^{2}}x=\dfrac{1-\cos 2x}{2}
So, we will get I=∫1−cos⁡(4πx)2dx \begin{aligned} & I=\int{\dfrac{1-\cos (4\pi x)}{2}dx} \\\ \end{aligned}
Now, we know that integration of dx is x and cosx is sinx.
So, using this we get,
⇒∫12dx−∫cos⁡(4πx)2dx ⇒12x−sin⁡4πx4×2×π \begin{aligned} & \Rightarrow \int{\dfrac{1}{2}}dx-\int{\dfrac{\cos (4\pi x)}{2}}dx \\\ & \Rightarrow \dfrac{1}{2}x-\dfrac{\sin 4\pi x}{4\times 2\times \pi } \\\ \end{aligned}
Now simplifying this after finding LCM we get,
I=4πx−sin⁡4πx8πxI=\dfrac{4\pi x-\sin 4\pi x}{8\pi x}
Hence, we got the integral for ∫(sin⁡(2πx))2dx\int{{{\left( \sin (2\pi x) \right)}^{2}}dx}.

Note: Remember the integration of six trigonometric functions directly in order to do your question quickly. Also, we must be aware of the differentiation to get the concepts of integration clearly and to use it wisely.