Question

How do I find the antiderivative of $y = \csc \left( x \right)\cot \left( x \right)$?

Answer 1

Hint: Here we are given to find the antiderivative which means integration of the terms. Here we are given that the function is $y = \csc \left( x \right)\cot \left( x \right)$ and we need to integrate it. So we substitute the value of $\csc x = \dfrac{1}{{\sin x}}$ and $\cot x = \dfrac{{\cos x}}{{\sin x}}$ and then we can simplify this integration.

Complete step by step solution:
Here we are given to find the antiderivative of $y = \csc \left( x \right)\cot \left( x \right)$ which means we need to do the integration of this term. So we can represent it as:
$\int {\csc \left( x \right)\cot \left( x \right)} .dx$$ - - - - (1)$
Now we need to simplify it by using the formula which says that:
$\csc x = \dfrac{1}{{\sin x}}$ and $\cot x = \dfrac{{\cos x}}{{\sin x}}$
Now we can substitute these formula values in the equation (1) and we will get as:
$\int {\dfrac{1}{{\sin x}}.\dfrac{{\cos x}}{{\sin x}}} .dx$
Now we know that when we multiply the two terms which are same then we can square that term and similar is the case here in the denominator where $\sin x$ is multiplied by itself and hence we can write it as:
$\int {\dfrac{{\cos x}}{{{{\sin }^2}x}}} .dx$$ - - - - (2)$
Here we do not have any direct formula to solve such a problem, so we can let $\sin x$ to be any variable and then its differentiation which is $\cos x$ which is given in the numerator. So we can easily simplify it and write it as:
Let $\sin x = t$
Differentiating both sides we get:
$\cos xdx = dt$
Now let us substitute this value which we have got after differentiation in equation (2) and we will get:
$\int {\dfrac{{\cos x}}{{{{\sin }^2}x}}} .dx$
$\int {\dfrac{{dt}}{{{t^2}}}} $
Now we know that $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c} $
Similarly we can say that:
$\int {\dfrac{{dt}}{{{t^2}}}} = \int {{t^{ - 2}}} dt$
So we will get the integration as:
$\int {\dfrac{{dt}}{{{t^2}}}} = \int {{t^{ - 2}}} dt = \dfrac{{{t^{ - 2 + 1}}}}{{ - 2 + 1}} = - \dfrac{1}{t}$
Now substituting the value of $t$ we get:

$\int {\csc \left( x \right)\cot \left( x \right)} .dx = - \dfrac{1}{{\sin x}} + c$

Note:
Here the student must know that the opposite of derivative is antiderivative and it is also termed as integration. If he does not know it then it would be difficult to proceed in the problem.