Question

How do you find the antiderivative of $\int{\left( {{\csc }^{3}}x \right)}dx$ ?

Answer 1

To solve the above integration i.e. $\int{\left( {{\csc }^{3}}x \right)}dx$, we are going to use integration by parts. The formula for integration by parts is as follows: $\int{udv}=uv-\int{vdu}$. Here u and v are two functions. Now, we will take $u=\csc x\And v=-\cot x$. Then we will take the derivative on both the sides. After doing that we will get $\int{udv}=\int{\left( {{\csc }^{3}}x \right)dx}$.

Complete step by step solution:
In the above problem, we are asked to find the integration of the following:
$\int{\left( {{\csc }^{3}}x \right)}dx$
Now, we are going to rewrite the above integration as follows:
$\int{\left( {{\csc }^{2}}x \right)\left( \csc x \right)}dx$
Now, we are going to apply integration by parts in the above integration by taking the two functions u and v as follows:
$\begin{aligned} & u=\csc x; \\\ & v=-\cot x \\\ \end{aligned}$
Now, we are going to differentiate on both the sides of this function as follows:
$\begin{aligned} & du=-\cot x\csc xdx; \\\ & dv={{\csc }^{2}}xdx \\\ \end{aligned}$
So, combining “u” and “dv” we get,
$\Rightarrow udv=\csc x\left( {{\csc }^{2}}x \right)dx$
Integrating on both the sides we get,
$\Rightarrow \int{udv}=\int{\csc x\left( {{\csc }^{2}}x \right)dx}$
$\Rightarrow \int{udv}=\int{\left( {{\csc }^{3}}x \right)dx}$
Now, applying integrating by parts on u and v we get,
$\Rightarrow \int{udv}=uv-\int{vdu}$
Substituting the value of $\int{udv}$ and “u and v” in the above integration we get,
$\Rightarrow \int{{{\csc }^{3}}x}=-\csc x\cot x-\int{{{\cot }^{2}}x\csc xdx}$
We know the trigonometric identity in ${{\cot }^{2}}x$ which is equal to:
$\Rightarrow {{\cot }^{2}}x={{\csc }^{2}}x-1$
Using the above relation in the above integration we get,

& \Rightarrow \int{{{\csc }^{3}}x}=-\csc x\cot x-\int{\left( {{\csc }^{2}}x-1 \right)\csc xdx} \\\ & \Rightarrow \int{{{\csc }^{3}}x}=-\csc x\cot x-\int{{{\csc }^{3}}xdx}+\int{\csc xdx} \\\ & \Rightarrow 2\int{{{\csc }^{3}}x}=-\csc x\cot x+\int{\csc xdx} \\\ \end{aligned}$$ We know the integration of $\csc x$ with respect to x which is shown below: $\int{\csc xdx}=-\ln \left( \left| \cot x+\csc x \right| \right)$ Using the above relation in the above integration we get, $$\Rightarrow 2\int{{{\csc }^{3}}x}=-\csc x\cot x-\ln \left( \left| \cot x+\csc x \right| \right)$$ Dividing 2 on both the sides we get, $$\Rightarrow \int{{{\csc }^{3}}x}=\dfrac{-\csc x\cot x-\ln \left( \left| \cot x+\csc x \right| \right)}{2}+C$$ In the above “C” is constant. **Hence, we have found the antiderivative of the above expression which is equal to: $$\Rightarrow \int{{{\csc }^{3}}x}=\dfrac{-\csc x\cot x-\ln \left( \left| \cot x+\csc x \right| \right)}{2}+C$$** **Note:** To solve the above integration we need to know the method of integration by parts. So, there is a trick to remember the formula for integration by parts which we are shown below: $\int{udv}=uv-\int{vdu}$ You can remember the “u and v” order from the alphabetical order that first “u” comes and then “v” will come. In the L.H.S, first “u” is there then “dv” will come. Then we multiply “u and v” together then we subtract integration of “vdu”. Now, you can remember “vdu” is the interchange of places of “udv”.