Question

What is the integral of \[\sin xdx\] from \[0\] to \[2\pi \]?

Answer 1

Hint: We are given a question to find the integral of sine function from \[0\] to \[2\pi \]. We will first write the given statement in the mathematical form, that is, \[\int\limits_{0}^{2\pi }{\sin xdx}\]. We will now integrate the sine function within the interval \[0\to 2\pi \]. We know that the integral of sine function is negative of cosine function, that is, \[\int{\sin x dx=-\cos x}\]. We will then apply the intervals and compute further to find the value of the integral. Hence, we will have the value of the integral.

Complete step-by-step solution:
According to the given question, we are given a trigonometric function which we have to integrate within the given interval.
The integral that we have is,
\[\int\limits_{0}^{2\pi }{\sin xdx}\]----(1)
We know that the integral of sine function is negative of cosine function, and it is represented as, \[\int{\sin x dx=-\cos x}\]
We will be applying this in our integral, we get the expression as,
\[\Rightarrow \left[ -\cos x \right]_{0}^{2\pi }\]----(2)
Now, we will apply the interval in the above expression, we get the new expression as,
\[\Rightarrow -\cos \left( 2\pi \right)-\left( -\cos 0 \right)\]-----(3)
We will now substitute the values of trigonometric functions present in the equation (3), we get the value as,
\[\Rightarrow -1-\left( -1 \right)\]
Opening up the brackets in the above equation, we get,
\[\Rightarrow -1+1\]
So, we have the value of the expression as,
\[\Rightarrow 0\]
Therefore, the value of \[\int\limits_{0}^{2\pi }{\sin xdx}=0\].

Note: The integration of sine function should not be confused with the differentiation of sine function. Both differ by a negative sign but that makes all of the difference. And while adding up the values make sure that negative sign cancels out wherever possible. Also, the values of trigonometric function at commonly used angles should be known beforehand and correctly too.