Question

Integration of $\sqrt {1 + 2\tan x(\sec x + \tan x)}$ wrt x is
(A) $\ln \left| {\sec x} \right| - \ln \left| {\sec x - \tan x} \right| + c$
(B) $\ln \left| {\sec x} \right| + \ln \left| {\sec x - \tan x} \right| + c$
(C) $2\ln \left| {\sec \dfrac{x}{2} + \tan \dfrac{x}{2}} \right| + c$
(D) $\ln \left| {1 + \tan x(\sec x + \tan x)} \right| + c$

Answer 1

Hint: We have to solve an integration firstly we put it equal to ${\text{I}}$. Ten we have to simplify the function in the integration by applying the identity. ${a^2} + 2ab + {b^2} = {(a + b)^2}$. After that we cancel the root by the square when the function gets simplified into its simplest form then we apply the formula of integration. Again we simplify it and get the answer.

Complete answer:
We have to integrate $\sqrt {1 + 2\tan x(\sec x + \tan x)} {\text{ w}}{\text{.r}}{\text{.t 'x'}}$
Let us take this integration equal to ${\text{I}}$.
$\Rightarrow$ ${\text{I = }}\int {\sqrt {1 + 2\tan x(\sec x + \tan x)} {\text{ dx}}} $
Now ${\text{I = is equal to }}{\sec ^2}x - {\tan ^2}x$
$\Rightarrow$${\text{I = }}\int {\sqrt {{{\sec }^2}x - {{\tan }^2}x + 2\sec x{\text{ }}\tan x + 2{{\tan }^2}x} {\text{ dx}}} $
Subtracting ${\tan ^2}x$ from $2{\tan ^2}x$, we get
${\text{I = }}\int {\sqrt {{{\sec }^2}x + 2\tan x{\text{ }}\sec x + {{\tan }^2}x} {\text{ dx}}} $
Here ${\sec ^2}x + 2\tan x{\text{ }}\sec x + {\tan ^2}x$ from identity ${a^2} + 2ab + {b^2}$ where $a$ is equal to $\sec x$, is equal to $\sec x,{\text{ }}b$ is equal to $\tan x$.
We know that ${a^2} + 2ab + {b^2} = {(a + b)^2}$
So we can write ${\sec ^2}x + 2\tan x{\text{ }}\sec x + {\tan ^2}x = {(\sec x + \tan x)^2}$
$\Rightarrow$ \[{\text{I = }}\int {\sqrt {{{(\sec x + \tan x)}^2}} {\text{ }}dx} \]
Cancelling square by square root
$\Rightarrow$ ${ I = }\int {\left| {\sec x + \tan x} \right|dx} $
Integrating $\sec x{\text{ and }}\tan x$ separately
$\Rightarrow$ \[{\text{I = }}\int {\left| {\sec x} \right|dx + \left| {\tan x} \right|} {\text{ }}dx\]
Now integration of $\left| {\sec x} \right|$ is equal to $\log \left| {\sec x + \tan x} \right|$ and integration of $\left| {\tan x} \right|$ is equal to $\log \left| {\sec x} \right|$.
Therefore \[{\text{I = log}}\left| {\sec x + \tan x} \right|{\text{ + log}}\left| {\sec x} \right| + c\]
As we have ${\sec ^2}x - {\tan ^2}x = 1$
Equating with identity ${a^2} - {b^2}$
as ${a^2} - {b^2} = (a + b)(a - b)$
So ${\sec ^2}x - {\tan ^2}x = (\sec x - \tan x)(\sec x + \tan x)$
Therefore $(\sec x - \tan x)(\sec x + \tan x) = 1$
$\Rightarrow$$(\sec x - \tan x) = \dfrac{1}{{(\sec x + \tan x)}}$
$\Rightarrow$ \[{\text{I = log}}\left| {\dfrac{1}{{\sec x + \tan x}}} \right|{\text{ + log}}\left| {\sec x} \right| + c\]
$\Rightarrow$ \[{\text{I = }} - \log \left| {\sec x + \tan x} \right|{\text{ + log}}\left| {\sec x} \right| + c\]

So option (A) is correct.

Note: When we take anti-derivative of a function then it represents the indefinite integral of the function. Indefinite integral has its symbol, a function and $dx$ in the end. It is an easy way of taking the antiderivative of a function. It is used to find the areas, volumes, central points and many useful things, but it is very simple to start with finding the area under the curve of function.