Question

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

Answer 1

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.