Question

What is the antiderivative of \[\dfrac{1}{{\sin x}}\]?

Answer 1

Hint: Integration is also known as the antiderivative in mathematics. We write the above-given function as \[f(x) = \dfrac{1}{{\sin x}}\]and then find its integral with respect to \[x\] which will give us the antiderivative of the function \[f(x)\].

Complete step-by-step solution:
We write the given trigonometric function as a function \[f(x)\]i.e. \[f(x) = \dfrac{1}{{\sin x}}\] and then we find its antiderivative or its integration with respect to \[x\].
That is \[\int {f(x)} {\kern 1pt} {\kern 1pt} dx\]\[ = \]\[\int {\dfrac{1}{{\sin x}}} \]\[dx\]
Now from our previous knowledge, we learned that \[\dfrac{1}{{\sin x}}\]is equal to \[\cos ec{\kern 1pt} {\kern 1pt} x\]
Therefore the integral changes to \[\int {f(x)} {\kern 1pt} {\kern 1pt} dx\]\[ = \]\[\int {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \cos ec{\kern 1pt} {\kern 1pt} x} \]\[dx\]
Then from the knowledge of integration, we know that the integration of \[{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \cos ec{\kern 1pt} {\kern 1pt} x\]is \[\ln (\cos ec{\kern 1pt} {\kern 1pt} x - \cot x)\]\[ + C\]
Therefore let us say that the integral of\[f{\kern 1pt} {\kern 1pt} (x)\]is \[F(x)\]and then we can write that
\[\int {f(x)} {\kern 1pt} {\kern 1pt} dx\]\[ = \]\[\int {\dfrac{1}{{\sin x}}} \]\[dx\]
\[ \Rightarrow \]\[F(x)\]\[ = \]\[\ln (\cos ec{\kern 1pt} {\kern 1pt} x - \cot x)\]\[ + C\]
Hence the antiderivative of the function \[f(x) = \dfrac{1}{{\sin x}}\]is equal to \[\ln (\cos ec{\kern 1pt} {\kern 1pt} x - \cot x)\]\[ + C\].
Additional information: The sine function i.e. \[\sin x\] and the cosecant function or \[{\kern 1pt} {\kern 1pt} {\kern 1pt} \cos ec{\kern 1pt} {\kern 1pt} x\] are reciprocal to each other. The former is the ratio of perpendicular by hypotenuse and the latter is its reciprocal that is hypotenuse by perpendicular. Though it is rarely needed to compute both of them together as once you find the value of one of the functions you simply take its reciprocal to get the other one. Also, a very trivial relationship that one can draw as an identity is that\[\sin x\]\[.\]\[\cos ecx\]\[ = \]\[1\]. The interval when both their graphs repeat their behavior or their period is\[2\pi \].

Note: It is very important that we know the behavior of all the trigonometric functions along with their graphs. Be cautious when finding the definite integral of some functions as they may not be continuous everywhere. Like the function \[\cos ecx\] is not defined at \[x = 0\] because as it is the reciprocal of \[\sin x\] when we take the reciprocal of \[0\] we find the quantity as not defined.