Question

How do you find the integral of $\sin \left( 2\pi t \right)dt$ ?

Answer 1

We first take $2\pi t$ to be another variable $x$ . This makes the integral a standard integral of the integration of $\sin x$ form. Integration of $\sin x$ gives $-\cos x$ . We then add the constant of integration $c$ . Finally, we replace $x$ with $2\pi t$

Complete step by step answer:
The given integrand is $\sin \left( 2\pi t \right)$ where the variable is $t$ . The integral becomes
$\int{\sin \left( 2\pi t \right)dt}....\text{integral1}$
Let us assume the entire term $2\pi t$ as another variable $x$ .
$\Rightarrow 2\pi t=x....equation1$
Differentiating both sides of the above equation and expressing it in differential forms, we get
$\Rightarrow 2\pi dt=dx$
Dividing both sides of the above equation by $2\pi$ we get,
$\Rightarrow dt=\dfrac{dx}{2\pi }$
Then, $\text{integral1}$ becomes,
$\Rightarrow \int{\sin \left( x \right)\dfrac{dx}{2\pi }}....\text{integral2}$
$\dfrac{1}{2\pi }$ being a constant term, we take it outside of the integral. The integral thus becomes,
$\Rightarrow \dfrac{1}{2\pi }\int{\sin \left( x \right)dx}$
We all know that the integration $\sin x$ gives us $-\cos x$ . The integral thus gets evaluated to
$\Rightarrow \dfrac{1}{2\pi }\left( -\cos x \right)$
As the integral given in the problem is an indefinite one, so we need to add a constant of integration $c$ . This constant of integration is added to include any constant term which can be present in the function, whose derivative we are integrating. For example, the derivative of ${{x}^{2}}$ and ${{x}^{2}}+c$ both are $2x$ . But, if we are given to integrate $2x$ , we have to consider the more general case which is ${{x}^{2}}+c$ . If the answer is ${{x}^{2}}$ then we can simply put $c$ as $0$ . Thus, always a constant of integration is taken into account.
So, we can write,
$\Rightarrow \dfrac{1}{2\pi }\left( -\cos x \right)+c$
We now replace $x$ with $2\pi t$ as the variable of the original integrand was $t$ and not $x$ . This means,
$\Rightarrow \dfrac{1}{2\pi }\left( -\cos 2\pi t \right)+c$
Therefore, we can conclude that $\int{\sin \left( 2\pi t \right)dt}$ evaluates to $\dfrac{1}{2\pi }\left( -\cos 2\pi t \right)+c$ .

Note:
If the integrand is of a complicated form, we can express some part of it as another variable. This will make the problem easier. But, finally we should convert back the assumed variables to the original ones. In case of definite integrals, we should put the limits carefully. For definite integrals, we must remember to include the constant of integration $c$ .