Question

Find m if\[\int {{{\sec }^2}x(1 - {{\cot }^2}x)dx = m\csc (2x) + C} \]

Answer 1

Hint: To find the unknown quantity m, simplify the integral on the left until no product of trigonometric functions are left. For this, open the brackets and then substitute for $\cot x$ using the quotient identities $\cot x = \dfrac{{\cos x}}{{\sin x}}$ and $\sec x = \dfrac{1}{{\cos x}}$. The final simplification will have an expression similar to $m\csc (2x) + C$. The value obtained after comparison will be the required answer.

Complete step by step answer:
Given that \[\int {{{\sec }^2}x(1 - {{\cot }^2}x)dx = m\csc (2x) + C} \]
Here, the value of the integral has a missing quantity m.
To find this missing quantity m, we will compute the integral given on the left hand side step-by-step.
We will be solving until we get $\csc (2x)$ in the solution.
So, we need to begin with the simplification of the expression in the left hand side of the given equation.
\[
  \int {{{\sec }^2}x(1 - {{\cot }^2}x)dx} \\
   = \int {({{\sec }^2}x - {{\sec }^2}x{{\cot }^2}x)dx} \\
 \]
Here, we use the quotient identities$\cot x = \dfrac{{\cos x}}{{\sin x}}$ and $\sec x = \dfrac{1}{{\cos x}}$
Note that on squaring both sides we get,
${\cot ^2}x = \dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}}$ and ${\sec ^2}x = \dfrac{1}{{{{\cos }^2}x}}$
Therefore, upon substitution we have the following:
\[
  \int {{{\sec }^2}x(1 - {{\cot }^2}x)dx} \\
   = \int {({{\sec }^2}x - \dfrac{1}{{{{\cos }^2}x}}\dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}})dx} \\
   = \int {({{\sec }^2}x - \dfrac{1}{{{{\sin }^2}x}})dx} \\
 \]
Here, we can use another quotient identity\[\csc x = \dfrac{1}{{\sin x}}\]
By squaring it on both the sides, we have\[{\csc ^2}x = \dfrac{1}{{{{\sin }^2}x}}\]
Therefore,
\[
  \int {{{\sec }^2}x(1 - {{\cot }^2}x)dx} \\
   = \int {({{\sec }^2}x - {{\csc }^2}x)dx} \\
 \]
We can separate the integrals using the following property of definite integration:\[\int {f(x) + g(x)dx = \int {f(x)dx + \int {g(x)dx} } } \]
\[
  \int {{{\sec }^2}x(1 - {{\cot }^2}x)dx} \\
   = \int {{{\sec }^2}xdx - \int {{{\csc }^2}x} dx} \\
 \]
We can now apply the integrals of the following trigonometric functions at this step directly.
(i)\[\smallint {\sec ^2}xdx = \tan x + C\]
(ii)\[\smallint {\csc ^2}xdx = - \cot x + C\]
Therefore,
\[
  \int {{{\sec }^2}x(1 - {{\cot }^2}x)dx} \\
   = \tan x - ( - \cot x) + C \\
   = \tan x + \cot x + C \\
 \]
We know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$ and $\cot x = \dfrac{{\cos x}}{{\sin x}}$
 So, we get
\[
  \int {{{\sec }^2}x(1 - {{\cot }^2}x)dx} \\
   = \dfrac{{\sin x}}{{\cos x}} + \dfrac{{\cos x}}{{\sin x}} + C \\
   = \dfrac{{{{\sin }^2}x + {{\cos }^2}x}}{{\cos x\sin x}} + C \\
 \]
We will use the Pythagorean identity ${\sin ^2}x + {\cos ^2}x = 1$ for further simplification.
This gives us
\[
  \int {{{\sec }^2}x(1 - {{\cot }^2}x)dx} \\
   = \dfrac{1}{{\cos x\sin x}} + C \\
 \]
We will multiply 2 in the numerator and the denominator so that we can use the double-angle identity $\sin 2x = 2\cos x\sin x$
\[
  \int {{{\sec }^2}x(1 - {{\cot }^2}x)dx} \\
   = \dfrac{2}{{2\cos x\sin x}} + C \\
   = \dfrac{2}{{\sin 2x}} + C \\
 \]
Using the quotient identity \[\csc x = \dfrac{1}{{\sin x}}\] again, we get
\[
  \int {{{\sec }^2}x(1 - {{\cot }^2}x)dx} \\
   = 2\csc (2x) + C \\
 \]

The value of the integral that we obtained needs to be compared with the expression $m\csc (2x) + C$ On comparing the two values of the integral, we get
$
  2\csc (2x) + C = m\csc (2x) + C \\
   \Rightarrow m = 2 \\
 $
Hence the value of m is 2.

Note: A common assumption students tend to make is that $\int {f(x)g(x)dx = \int {f(x)dx\int {g(x)dx} } } $
This is not true in general and hence trying to solve using this approach will most likely lead you to a wrong answer.
It is best to try to simplify any product of trigonometric functions given in the integral into the sum or difference of trigonometric functions to find the solution in an easier way.