Question

What is antiderivative of ${{\csc }^{2}}x?$

Answer 1

Hint: To solve this question we should know the concept of integration. Integration of certain functions can also be written as the antiderivative of that function. This also requires the knowledge of the trigonometric function. Start by considering $\csc x=\dfrac{1}{\sin x}$ and then find its integral by substituting $u=\cot x$ .

Complete step by step solution:
Starting the solution with the trigonometric formula to convert $\csc x$ or $\text{cosec} x$ in terms of $\sin x$, we get
$\csc x=\dfrac{1}{\sin x}$,
Now solving the equation,
$\int{{{\csc }^{2}}x}dx=\int{\dfrac{1}{{{\sin }^{2}}x}}dx$
Let us consider $u=\cot x$
Further changing $\cot x$ in terms of $\sin x$ and $\cos x$, we get
 $\cot x=\dfrac{\cos x}{\sin x}$
So further we get,
 $u=\cot x=\dfrac{\cos x}{\sin x}$
On differentiating $u$ with respect to $x$
$\dfrac{du}{dx}=\dfrac{d(\cot x)}{dx}$
On changing $\cot x$ in terms of $\sin x$ and $\cos x$ ,
$\Rightarrow \dfrac{d\left( \dfrac{\cos x}{\sin x} \right)}{dx}$
The formula used here for differentiation of $u$ and $v$ , when both are the function in terms of $x$ . Then differentiation of their division of $u$ and $v$is \[\dfrac{d\left( \dfrac{u}{v} \right)}{dx}=\dfrac{v\left( \dfrac{du}{dx} \right)-u\left( \dfrac{dv}{dx} \right)}{{{v}^{2}}}\]. Applying same formula in the above we get,
$\Rightarrow \dfrac{d\left( \dfrac{\cos x}{\sin x} \right)}{dx}=\dfrac{\sin x\left( \dfrac{d\cos x}{dx} \right)-\cos x\left( \dfrac{d\sin x}{dx} \right)}{\sin {{x}^{2}}}$
Differentiation of $\sin x$ is $\cos x$ and differentiation of $\cos x$ is $-\sin x$ , on applying these formula in the above equation, we get
\[\Rightarrow \dfrac{du}{dx}=\dfrac{\sin x(-\sin x)-\cos x(\cos x)}{{{\sin }^{2}}x}\]
On multiplying the terms we get,
$\Rightarrow \dfrac{du}{dx}=\dfrac{-({{\sin }^{2}}x+{{\cos }^{2}}x)}{{{\sin }^{2}}x}$
Now we need to apply the identity in the formula
$\Rightarrow \dfrac{du}{dx}=\dfrac{-1}{{{\sin }^{2}}x}$
We know that $\dfrac{1}{\sin x}=\cos ecx$ , squaring both the function it becomes $\dfrac{1}{{{\sin }^{2}}x}=\cos e{{c}^{2}}x$ so applying same in the formula, we get the value as:
$\Rightarrow \dfrac{du}{dx}=-\cos e{{c}^{2}}x$
$\Rightarrow -du=\cos e{{c}^{2}}xdx={{\csc }^{2}}xdx$
On substituting the value $du$ in terms of ${{\csc }^{2}}xdx$ , we get
$\int{{{\csc }^{2}}x}=-\int{du}$
On integrating, the above integral we get,
$\Rightarrow -\int{du}=-u+c$
Substituting the value of $u$ in the above equation as $\cot x$ , we get
$\Rightarrow -\cot x+c$

$\therefore $ The antiderivative of the ${{\csc }^{2}}x$ is $-\cot x+c$.

Note: We can check whether the antiderivative is right or not. We can differentiate the result , so for differentiating $-\cot x+c$ with respect to $x$ these are the process that need to be undertaken,
$\dfrac{d\left( -\cot x+c \right)}{dx}$
$\Rightarrow \dfrac{d(-\cot x)}{dx}+\dfrac{dc}{dx}$
Differentiation of $-\cot x$ is $\cos e{{c}^{2}}x$ and differentiation of a constant $c$ is $0$.
$\Rightarrow \cos e{{c}^{2}}x+0$
Since, the derivative of the answer is the same as that of the question so the antiderivative of $\cos e{{c}^{2}}x$ is correct.