Question

Find the value of the integral function $\int {\dfrac{{dx}}{{1 + \cos x + \sin x}}} $.
A. $\dfrac{1}{4}\log \left| {1 + \tan \dfrac{x}{2}} \right| + C$
B. $\dfrac{1}{2}\log \left| {1 + \tan \dfrac{x}{2}} \right| + C$
C. $\log \left| {1 + \tan \dfrac{x}{2}} \right| + C$
D. $\dfrac{1}{2}\log \left| {1 - \tan \dfrac{x}{2}} \right| + C$

Answer 1

Hint: In this case, we must apply the $\sin x$ and $\cos x$ trigonometric formulas in the form of $\tan x$. By substituting $x$ with $\dfrac{x}{2}$ and modifying the changes to the trigonometric identity and the $\cos 2x$ and $\sin 2x$ formulas. Finally, we must use the basic integration formula to determine the value.

Formula Used:
The trigonometric formula are given as
1. $\cos 2x = \dfrac{{1 - {{\tan}^2}x}}{{1 + {{\tan}^2}x}}$
2. $\sin 2x = \dfrac{{2\tan x}}{{1 + {{\tan}^2}x}}$
3. $\int {\dfrac{{dt}}{t} = \log \left| t \right| + c} $
4. $1 + {\tan ^2}x = {\sec ^2}x$

Complete step by step solution:
Given that the integral is $\int {\dfrac{{dx}}{{1 + \cos x + \sin x}}} $.
Let us consider $I = \int {\dfrac{{dx}}{{1 + \cos x + \sin x}}} $ —(1)
As we know that, $\cos 2x = \dfrac{{1 - {{\tan}^2}x}}{{1 + {{\tan}^2}x}}$ and we can express $\cos x$ as
$\cos x = \dfrac{{1 - {{\tan}^2}\dfrac{x}{2}}}{{1 + {{\tan}^2}\dfrac{x}{2}}}$
Also, we know that $\sin 2x = \dfrac{{2\tan x}}{{1 + {{\tan}^2}x}}$ and we an express $\sin x$ as $\sin x = \dfrac{{2\tan \dfrac{x}{2}}}{{1 + {{\tan}^2}\dfrac{x}{2}}}$.
Now, we will substitute these values in equation (1), we get
$I = \int {\dfrac{{dx}}{{1 + \dfrac{{1 - {{\tan}^2}\dfrac{x}{2}}}{{1 + {{\tan}^2}\dfrac{x}{2}}} + \dfrac{{2\tan \dfrac{x}{2}}}{{1 + {{\tan}^2}\dfrac{x}{2}}}}}} $
Further, we will simplify the above expression, we get
$I = \int {\dfrac{{dx}}{{\dfrac{{1 + {{\tan}^2}\dfrac{x}{2} + 1 - {{\tan}^2}\dfrac{x}{2} + 2\tan \dfrac{x}{2}}}{{1 + {{\tan}^2}\dfrac{x}{2}}}}}} \\I = \int {\dfrac{{\left( {1 + {{\tan}^2}\dfrac{x}{2}} \right)dx}}{{2 + 2\tan \dfrac{x}{2}}}}$
Furthermore, we will take out $2$ as common from the denominator, we get
$I = \int {\dfrac{{\left( {1 + {{\tan}^2}\dfrac{x}{2}} \right)dx}}{{2\left( {1 + \tan \dfrac{x}{2}} \right)}}} $
Now, we will use the formula $1 + {\tan ^2}x = {\sec ^2}x$ and express
$1 + {\tan ^2}\dfrac{x}{2} = {\sec ^2}\dfrac{x}{2}$, we get
$I = \int {\dfrac{{{{\sec}^2}\dfrac{x}{2}dx}}{{2\left( {1 + \tan \dfrac{x}{2}} \right)}}} $ —(2)
Further, let us assume $\tan \dfrac{x}{2} = t$ and the derivative of $\tan \dfrac{x}{2}$ is $\dfrac{1}{2}{\sec ^2}\dfrac{x}{2}dx = dt$.
Now, we will substitute the values in equation (2), we get
$I = \int {\dfrac{{dt}}{{1 + t}}} $
Furthermore, we will apply the formula $\int {\dfrac{{dt}}{t} = \log \left| t \right| + c} $, we get
$I = \log \left| {1 + t} \right| + C$
Again, we will substitute $t = \tan \dfrac{x}{2}$, we get
$I = \log \left| {1 + \tan \dfrac{x}{2}} \right| + C$

Option ‘C’ is correct

Note: In these types of questions, we should remember the conversions of trigonometric functions and trigonometric identity also. We should be careful while adjusting the formulas of trigonometric functions so that no error will occur and also know the basic differentiation formulas and integration formulas.