Question

Evaluate \[\int {\dfrac{{{{\sec }^2}x}}{{3 + {{\tan }^2}x}}dx} \].

Answer 1

Hint: Here we will use the basic integration formula which states that integration of any variable \[x\] will be equals to as below:
\[\int {\dfrac{1}{{1 + {x^2}}}dx = {{\tan }^{ - 1}}x + {\text{C}}} \] , where \[x\] is any variable and \[{\text{C}}\] is an arbitrary constant.

Complete step-by-step solution:
Step 1: In the given expression \[I = \int {\dfrac{{{{\sec }^2}x}}{{3 + {{\tan }^2}x}}dx} \] , let \[\tan x = t\]. Therefore, \[{\sec ^2}xdx = dt\].
By substituting the value of \[{\sec ^2}xdx = dt\] in the given expression \[I = \int {\dfrac{{{{\sec }^2}x}}{{3 + {{\tan }^2}x}}dx} \] , we get:
\[ \Rightarrow I = \int {\dfrac{{dt}}{{3 + {t^2}}}} \]
By taking \[3\] out of the integral and writing \[3 = {\left( {\sqrt 3 } \right)^2}\] in the expression \[I = \int {\dfrac{{dt}}{{3 + {t^2}}}} \] , we get:
\[ \Rightarrow I = \dfrac{1}{{{{\left( {\sqrt 3 } \right)}^2}}}\int {\dfrac{{dt}}{{1 + \dfrac{{{t^2}}}{{{{\left( {\sqrt 3 } \right)}^2}}}}}} \] ………… (1)
Step 2: In the expression (1), by bringing \[\dfrac{1}{{\sqrt 3 }}\] inside the derivative symbol, we get:
\[ \Rightarrow I = \dfrac{1}{{\sqrt 3 }}\int {\dfrac{{d\dfrac{t}{{\sqrt 3 }}}}{{1 + {{\left( {\dfrac{t}{{\sqrt 3 }}} \right)}^2}}}} \]
Now, by using the formula \[\int {\dfrac{1}{{1 + {x^2}}}dx = {{\tan }^{ - 1}}x + {\text{C}}} \] in the above expression, we get:
\[ \Rightarrow I = \dfrac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\dfrac{t}{{\sqrt 3 }} + {\text{C}}\] , where \[{\text{C}}\] is an arbitrary constant.
Step 3: By substituting the value of \[\tan x = t\] in the expression \[I = \dfrac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\dfrac{t}{{\sqrt 3 }} + {\text{C}}\] , we get:
\[ \Rightarrow I = \dfrac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\dfrac{{\tan x}}{{\sqrt 3 }}} \right) + {\text{C}}\]

The value of the integral \[\int {\dfrac{{{{\sec }^2}x}}{{3 + {{\tan }^2}x}}dx} = \dfrac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\dfrac{{\tan x}}{{\sqrt 3 }}} \right) + {\text{C}}\]

Note: Students needs to remember that \[\int {{{\sec }^2}xdx} = \tan x + {\text{C}}\] , the explanation is given below for better understanding:
Let \[I = \int {{{\sec }^2}xdx} \]
By multiplying and dividing the expression \[\int {{{\sec }^2}xdx} \] with \[\tan x\] , we get:
\[ \Rightarrow I = \int {\dfrac{{{{\sec }^2}x\tan x}}{{\tan x}}dx} \]
Let, \[\sec x = t\] and by taking the derivative of it, we get \[\sec x\tan xdx = dt\].
Also, we know that, \[\tan x = \sqrt {{{\sec }^2}x - 1} \]. So, by putting these values in the expression \[ \Rightarrow I = \int {\dfrac{{{{\sec }^2}x\tan x}}{{\tan x}}dx} \] , we get:
\[ \Rightarrow I = \int {\dfrac{t}{{\sqrt {{t^2} - 1} }}dt} \]
Assume that \[{t^2} - 1 = u\], so by taking the derivative of it we get:
\[ \Rightarrow \dfrac{d}{{dx}}\left( {{t^2} - 1} \right) = du\]
By simplifying the LHS side, we get:
\[ \Rightarrow 2tdt = du\]
By doing the integration in the above expression, we get:
\[ \Rightarrow I = \dfrac{1}{2}\int {{u^{ - \dfrac{1}{2}}}du} \]
By simplifying the terms in the RHS side, we get:
\[ \Rightarrow I = \left( {\dfrac{1}{2}} \right)\dfrac{{\sqrt u }}{{\left( {\dfrac{1}{2}} \right)}}\]
\[ \Rightarrow I = \sqrt u \]
By substituting the value of \[{t^2} - 1 = u\]in the above expression \[I = \sqrt u \], we get:
\[ \Rightarrow I = \sqrt {{t^2} - 1} \]
Now, by substituting the value of \[\sec x = t\] in the above expression \[I = \sqrt {{t^2} - 1} \], we get:
\[ \Rightarrow I = \sqrt {{{\sec }^2}x - 1} \]
But we know that \[{\sec ^2}x - 1 = \tan x\], so by substituting this value in the above expression we get:
\[ \Rightarrow I = \sqrt {{{\tan }^2}x} \]
\[ \Rightarrow I = \tan x + {\text{C}}\] , where \[{\text{C}}\] is an arbitrary constant.