Question

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

Answer 1

: In this question, we have to find the antiderivative of an equation. Antiderivative means the inverse of the derivation, which is also known as integration. The equation given to us consists of the trigonometric functions; therefore, we will solve this problem using the integration of trigonometric functions. We start solving this problem by converting the trigonometric function into sin and cos functions. After that, we will make the necessary calculations and let the square root of the denominator be equal to t, that is $\sin x=t$ . Then, we will take the derivative of the substitute function and solve it further. Thus, we again use the integration formula $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}$ in the equation and again substitute the value of t in the equation, to get the required solution to the problem.

Complete answer:
According to the question, we have to find the antiderivative of the trigonometric function.
Thus, we will apply the formula of the integration of the trigonometric function.
The problem we have to solve is $y=\csc (x)\cot (x)$ .
Thus, we first change the given trigonometric functions into sin and cos function, that is $\cos ecx=\dfrac{1}{\sin x}$ and $\cot x=\dfrac{\cos x}{\sin x}$ , so we will substitute in the equation (1), we get
$\Rightarrow y=\dfrac{1}{\sin x}.\left( \dfrac{\cos x}{\sin x} \right)$
On further solving, we get
$\Rightarrow y=\left( \dfrac{\cos x}{{{\sin }^{2}}x} \right)$
Now, we will take the integration sign on the right-hand side of the above equation, we get
$\Rightarrow y=\int{\left( \dfrac{\cos x}{{{\sin }^{2}}x} \right)dx}$ ----------- (2)
Now, let the square root of the denominator in the equation (2) equals t, that is
$\sin x=t$ ---------- (3)
Now, take the derivative of the above equation, we get
$\cos xdx=dt$
$\Rightarrow dx=\dfrac{dt}{\cos x}$ --------- (4)
Thus, substituting the value of equation (3) and (4) in equation (2), we get
$\Rightarrow y=\int{\left( \dfrac{\cos x}{{{t}^{2}}} \right)dt.\dfrac{1}{\cos x}}$
Therefore, both the cosx in the numerator and the denominator cancel out each other, thus we get

& \Rightarrow y=\int{\left( \dfrac{1}{{{t}^{2}}} \right)dt} \\\ & \Rightarrow y=\int{{{t}^{-2}}.dt} \\\ \end{aligned}$$ Now, we will apply the formula $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}$ in the above equation, we get $\Rightarrow y=\dfrac{{{t}^{(-2+1)}}}{-2+1}$ On further simplification, we get $\Rightarrow y=\dfrac{{{t}^{-1}}}{-1}$ $\Rightarrow y=-\dfrac{1}{t}$ Thus, now, we will substitute the value of t from equation (3) in the above equation, we get $y=\dfrac{-1}{\sin x}$ Therefore, for the equation $y=\csc (x)\cot (x)$ , the value of its antiderivative is $-\dfrac{1}{\sin x}$. **Note:** While solving this problem, keep in mind the formulas you are using to avoid mathematical errors and do the step-by-step calculations to get an accurate answer. One of the alternatives methods to solve this problem is by using the by-parts method, where let $\csc (x)=u$ and $\cot (x)=v$, put these two equations in the formula, and get the required solution to the problem.