Question

What is the integral of sec(x)?

Answer 1

Hint: Here, we multiply the given function by \[\dfrac{{\sec x + \tan x}}{{\sec x + \tan x}}\] which is same as multiplying by\[1\]. Also we, will use logarithmic derivative which is\[\dfrac{{du}}{u} = \ln |u|\]. We will also use \[\cos x = \dfrac{1}{{\sec x}}\]. And we have trigonometry identity used in our solution, which is, \[{\sin ^2}x + {\cos ^2}x = 1\].

Complete answer: Here, given that,
\[\int {\sec (x)dx} \]
Multiplying sec(x) by\[\dfrac{{\sec x + \tan x}}{{\sec x + \tan x}}\], which is really the same as multiplying by 1.
Thus, we have,
\[ = \int {\sec x(\dfrac{{\sec x + \tan x}}{{\sec x + \tan x}})} dx\]
Simplify the given expression, we get,
\[ = \int {\dfrac{{\sec x(\sec x + \tan x)}}{{(\sec x + \tan x)}}dx} \]
Removing the brackets, we get,
\[ = \int {\dfrac{{{{\sec }^2}x + \sec x\tan x}}{{\sec x + \tan x}}} dx\]
Let us substitute the value of \[\sec x + \tan x\]as u.
i.e. \[u = \sec x + \tan x\]
Then,
\[du = (\sec x\tan x + {\sec ^2}x)dx\]
Rearranging the given expression, we get,
\[du = ({\sec ^2}x + \sec x\tan x)dx\]
Thus,
\[\dfrac{{du}}{u} = \ln |u|\]
This, is called logarithmic derivative:
Now, integrating both the sides, we get,
\[\int {\dfrac{{du}}{u} = \ln |u| + C} \]
Here, replace the u with x.
Now substituting the values of u in the above integral function, we get,
 \[\int {\sec xdx = \ln (\sec x + \tan x) + C} \]
Other methods to solve this integration, we can do the following.
\[\int {\dfrac{1}{{\cos x}}dx} \]
Multiply with cos x in both numerator and denominator, we get,
\[
   = \int {\dfrac{1}{{\cos x}} \times \dfrac{{\cos x}}{{\cos x}}dx} \\
   = \int {\dfrac{{\cos x}}{{{{\cos }^2}x}}dx} \\
 \]
Since \[{\cos ^2}x = 1 - {\sin ^2}x\], we get,
\[ = \int {\dfrac{{\cos x}}{{1 - {{\sin }^2}x}}dx} \]
Substituting the values of y, we get,
\[ = \int {\dfrac{1}{{1 - {y^2}}}dx} \]
Also, \[1 - {y^2} = (1 + y)(1 - y)\]
\[
   = \int {\dfrac{1}{{(1 + y)(1 - y)}}dx} \\
   = \int {\dfrac{1}{2}(\dfrac{1}{{1 + y}} + \dfrac{1}{{1 - y}})dx} \\
   = \dfrac{1}{2}\int {(\dfrac{1}{{1 + y}} + \dfrac{1}{{1 - y}})dx} \\
 \]
Rearranging the above integration, we get,
\[
   = \dfrac{1}{2}\int {(\dfrac{1}{{y + 1}} - \dfrac{1}{{y - 1}})dx} \\
   = \dfrac{1}{2}[\ln |y + 1| - \ln |y - 1|] + C \\
    \\
 \]
This is called logarithmic derivative.
\[
   = \dfrac{1}{2}[\dfrac{{\ln |y + 1|}}{{\ln |y - 1|}}] + C \\
   = \dfrac{1}{2}\ln |\dfrac{{y + 1}}{{y - 1}}| + C \\
 \]
Now, substituting the values of y, we get,
\[ = \dfrac{1}{2}\ln |\dfrac{{\sin x + 1}}{{\sin x - 1}}| + C\]
Further to check if the answer is correct or not, we will do as below.
\[\dfrac{{\sin x + 1}}{{\sin x - 1}}\]
Multiply the above expression with \[\dfrac{{\sin x + 1}}{{\sin x - 1}}\], we get,
=\[\dfrac{{\sin x + 1}}{{\sin x - 1}} \times \dfrac{{\sin x + 1}}{{\sin x + 1}}\]
\[ = \dfrac{{{{(\sin x + 1)}^2}}}{{{{\sin }^2}x - 1}}\]
Since \[{\cos ^2}x = 1 - {\sin ^2}x\], so using this we get,
\[ = \dfrac{{{{(\sin x + 1)}^2}}}{{ - {{\cos }^2}x}}\] --- (i)
Now,
\[\dfrac{1}{2}\ln |\dfrac{{\sin x + 1}}{{\sin x - 1}}| + C\]
Substituting the (i), we get,
\[ = \dfrac{1}{2}\ln |\dfrac{{{{(\sin x + 1)}^2}}}{{ - {{\cos }^2}x}}| + C\]
Subtract sign is removed because of the mode used.
\[ = \dfrac{1}{2}\ln |\dfrac{{{{(\sin x + 1)}^2}}}{{{{\cos }^2}x}}| + C\]
Simplify this we get,
\[
   = \ln |\dfrac{{\sin x + 1}}{{\cos x}}| + C \\
   = \ln |\dfrac{{\sin x}}{{\cos x}} + \dfrac{1}{{\cos x}}| + C \\
 \]
Here, \[\dfrac{{\sin x}}{{\cos x}} = \tan x\]and \[\dfrac{1}{{\cos x}} = \sec x\]
Thus, substituting these values, we get,
\[ = \ln |\tan x + \sec x| + C\]

Note:
By taking the derivative of exactly the right function and looking at the results in the right way we got the formula we needed. We can solve this by using partial functions too. Then we need to use \[y = \sin x\] and \[dy = \cos xdx\]. One should know derivation before solving integration questions.