Question

How do you find the antiderivative of $ \int{{{\sec }^{2}}x{{\csc }^{2}}xdx} $ ?

Answer 1

Hint: In the problem we have the trigonometric ratios $ \sec x $ and $ \csc x $ . For integration we will use the trigonometric identity which is $ {{\csc }^{2}}x-{{\cot }^{2}}x=1 $ from this identity we will substitute $ {{\csc }^{2}}x=1+{{\cot }^{2}}x $ in the given integration and simplify the equation by using distribution law of multiplication. Now we will convert the possible trigonometric ratios into simple trigonometric ratios like $ \sin x $ , $ \cos x $ . After doing all the steps then we will get that the given equation which is in the form of $ \int{uvdx} $ is converted into $ \int{\left( u+v \right)dx} $ . Now it is easier to apply the integration and using the proper integration formulas we will get the required result.

Complete step by step answer:
Given that, $ \int{{{\sec }^{2}}x{{\csc }^{2}}xdx} $ .
From the trigonometric identity $ {{\csc }^{2}}x-{{\cot }^{2}}x=1 $ , substituting $ {{\csc }^{2}}x=1+{{\cot }^{2}}x $ in the above equation, then we will get
 $ \int{{{\sec }^{2}}x{{\csc }^{2}}xdx}=\int{{{\sec }^{2}}x\left( 1+{{\cot }^{2}}x \right)}dx $
Applying distribution law of multiplication, then we will get
 $ \Rightarrow \int{{{\sec }^{2}}x{{\csc }^{2}}xdx}=\int{\left( {{\sec }^{2}}x+{{\sec }^{2}}x.{{\cot }^{2}}x \right)dx} $
We know that $ \sec x=\dfrac{1}{\cos x} $ , $ \cot x=\dfrac{\cos x}{\sin x} $ . From this value the above equation is modified as
 $ \Rightarrow \int{{{\sec }^{2}}x{{\csc }^{2}}xdx}=\int{\left( {{\sec }^{2}}x+\dfrac{1}{{{\cos }^{2}}x}\times \dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x} \right)dx} $
Simplifying the above equation, then we will get
 $ \Rightarrow \int{{{\sec }^{2}}x{{\csc }^{2}}xdx}=\int{\left( {{\sec }^{2}}x+\dfrac{1}{{{\sin }^{2}}x} \right)dx} $
Using the trigonometric formula $ \csc x=\dfrac{1}{\sin x} $ , then we will get
 $ \Rightarrow \int{{{\sec }^{2}}x{{\csc }^{2}}x}dx=\int{\left( {{\sec }^{2}}x+{{\csc }^{2}}x \right)dx} $
Applying integration for each term individually, then we will get
 $ \Rightarrow \int{{{\sec }^{2}}x{{\csc }^{2}}xdx}=\int{{{\sec }^{2}}x}dx+\int{{{\csc }^{2}}x}dx $
We know that $ \int{{{\sec }^{2}}xdx}=\tan x+C $ , $ \int{{{\csc }^{2}}xdx}=-\cot x+C $ , then we will get
$ \Rightarrow \int{{{\sec }^{2}}x{{\csc }^{2}}xdx}=\tan x-\cot x+C $.

Note:
 We can also solve the above problem without using the trigonometric identity $ {{\csc }^{2}}x-{{\cot }^{2}}x=1 $ .
We know that $ \sec x=\dfrac{1}{\cos x} $ , $ \csc x=\dfrac{1}{\sin x} $ , then the given equation is modified as
 $ \int{{{\sec }^{2}}x{{\csc }^{2}}x}dx=\int{\dfrac{1}{{{\cos }^{2}}x}.\dfrac{1}{{{\sin }^{2}}x}}dx $
We have the trigonometric identity $ {{\sin }^{2}}x+{{\cos }^{2}}x=1 $ , then we will get
 $ \int{{{\sec }^{2}}x{{\csc }^{2}}xdx}=\int{\dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{{{\sin }^{2}}x.{{\cos }^{2}}x}dx} $
Simplifying the above equation, then we will have
 $ \begin{align}
  & \Rightarrow \int{{{\sec }^{2}}x{{\csc }^{2}}xdx}=\int{\left( \dfrac{{{\sin }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}+\dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x} \right)dx} \\
 & \Rightarrow \int{{{\sec }^{2}}x{{\csc }^{2}}xdx}=\int{\left( \dfrac{1}{{{\cos }^{2}}x}+\dfrac{1}{{{\sin }^{2}}x} \right)dx} \\
 & \Rightarrow \int{{{\sec }^{2}}x{{\csc }^{2}}xdx}=\int{\left( {{\sec }^{2}}x+{{\csc }^{2}}x \right)dx} \\
\end{align} $
Applying the integration to each term individually and using the know formulas we will get the result as
 $ \Rightarrow \int{{{\sec }^{2}}x{{\csc }^{2}}xdx}=\tan x-\cot x+C $
From both the methods we got the same result.