Question

Evaluate $\int {\dfrac{{1 - \cos x}}{{\cos x\left( {1 + \cos x} \right)}}dx}$ ?
A. $\log \left( {\sec x + \tan x} \right) - 2\tan \left( {\dfrac{x}{2}} \right) + C$
B. $\log \left( {\sec x + \tan x} \right) + 2\tan \left( {\dfrac{x}{2}} \right) + C$
C. $\log \left( {\sec x - \tan x} \right) - 2\tan \left( {\dfrac{x}{2}} \right) + C$
D. $\log \left( {\sec x - \tan x} \right) + 2\tan \left( {\dfrac{x}{2}} \right) + C$

Answer 1

This question is just to check how we can use the integration and derivative identities. We will try to shift the cos function into sec and tan function. First we will separate the integration terms above. Then we will use the combination according to the way to proceed towards the solution. Meanwhile, we will use the formulas given below.

Formula used:
$\int {{{\sec }^2}\left( {\dfrac{x}{2}} \right)dx} = \tan \left( {\dfrac{x}{2}} \right)$
$\Rightarrow \dfrac{d}{{dx}}\left( {\tan x} \right) = {\sec ^2}x$
$\Rightarrow \dfrac{d}{{dx}}\left( {\sec x} \right) = \sec x.\tan x$

Complete step by step answer:
We are given with an integrative function,
$\int {\dfrac{{1 - \cos x}}{{\cos x\left( {1 + \cos x} \right)}}dx}$
We will separate the terms first,
$\int {\dfrac{{1 - \cos x}}{{\cos x\left( {1 + \cos x} \right)}}dx} = \int {\dfrac{{dx}}{{\cos x\left( {1 + \cos x} \right)}} - \int {\dfrac{{\cos x}}{{\cos x\left( {1 + \cos x} \right)}}} dx}$
Cancelling cos function from the second term we get,
$\int {\dfrac{{1 - \cos x}}{{\cos x\left( {1 + \cos x} \right)}}dx} = \int {\dfrac{{dx}}{{\cos x\left( {1 + \cos x} \right)}} - \int {\dfrac{{dx}}{{\left( {1 + \cos x} \right)}}} }$

We can rearrange the first term as,
$\int {\dfrac{{1 - \cos x}}{{\cos x\left( {1 + \cos x} \right)}}dx}= \int {\left( {\dfrac{1}{{\cos x}} - \dfrac{1}{{1 + \cos x}}} \right)dx - } \int {\dfrac{{dx}}{{\left( {1 + \cos x} \right)}}}$
Taking LCM of the first two terms by cross multiplying we get,
$\int {\dfrac{{1 - \cos x}}{{\cos x\left( {1 + \cos x} \right)}}dx} = \int {\dfrac{{1 + \cos x - \cos x}}{{\cos x\left( {1 + \cos x} \right)}}dx - \int {\dfrac{{dx}}{{\left( {1 + \cos x} \right)}}} }$
Cancelling the common cos term from the numerator,
$\int {\dfrac{{1 - \cos x}}{{\cos x\left( {1 + \cos x} \right)}}dx} = \int {\dfrac{1}{{\cos x\left( {1 + \cos x} \right)}}dx - \int {\dfrac{{dx}}{{\left( {1 + \cos x} \right)}}} }$

Now we again separate the integrates,
$\int {\dfrac{{1 - \cos x}}{{\cos x\left( {1 + \cos x} \right)}}dx} = \int {\dfrac{1}{{\cos x}}dx - \int {\dfrac{1}{{1 + \cos x}}dx - \int {\dfrac{1}{{\left( {1 + \cos x} \right)}}dx} } }$
Adding the last two integrates since they are same we get,
$\int {\dfrac{{1 - \cos x}}{{\cos x\left( {1 + \cos x} \right)}}dx}= \int {\dfrac{1}{{\cos x}}dx - \int {\dfrac{2}{{1 + \cos x}}dx} }$
We know that $\Rightarrow \cos x = 2{\cos ^2}\left( {\dfrac{x}{2}} \right) - 1$ and the reciprocal of cos is sec. so changing these we can write,
$\int {\dfrac{{1 - \cos x}}{{\cos x\left( {1 + \cos x} \right)}}dx}= \int {\sec xdx - \int {\dfrac{2}{{1 + 2{{\cos }^2}\left( {\dfrac{x}{2}} \right) - 1}}} } dx$
Cancelling the 1,
$\int {\dfrac{{1 - \cos x}}{{\cos x\left( {1 + \cos x} \right)}}dx} = \int {\sec xdx - \int {\dfrac{2}{{2{{\cos }^2}\left( {\dfrac{x}{2}} \right)}}dx} }$

Now cancelling 2 and again taking the reciprocal of cos,
$\int {\dfrac{{1 - \cos x}}{{\cos x\left( {1 + \cos x} \right)}}dx} = \int {\sec xdx - \int {{{\sec }^2}\left( {\dfrac{x}{2}} \right)dx} }$
For the second integral we know the perfect integration $\Rightarrow \int {{{\sec }^2}\left( {\dfrac{x}{2}} \right)dx} = \tan \left( {\dfrac{x}{2}} \right) + C$
So ,
$\int {\dfrac{{1 - \cos x}}{{\cos x\left( {1 + \cos x} \right)}}dx}= \int {\sec xdx - } \tan \left( {\dfrac{x}{2}} \right) + C$
Now for the remaining integral we will try to use substitution but by adding a term in numerator and denominator,
$\int {\dfrac{{1 - \cos x}}{{\cos x\left( {1 + \cos x} \right)}}dx}= \int {\dfrac{{\sec x\left( {\sec x + \tan x} \right)}}{{\sec x + \tan x}}} dx - \tan \left( {\dfrac{x}{2}} \right) + C$
On multiplying with bracket,
$\int {\dfrac{{1 - \cos x}}{{\cos x\left( {1 + \cos x} \right)}}dx} = \int {\dfrac{{{{\sec }^2}x + \sec x.\tan x}}{{\sec x + \tan x}}} dx - \tan \left( {\dfrac{x}{2}} \right) + C$

Now substitute,
$\Rightarrow \sec x + \tan x = m$
Taking the derivative, from the formula mentioned above we get,
$\left( {\sec x.\tan x + {{\sec }^2}x} \right)dx = dm$
Now substitute this is the integral above,
$\int {\dfrac{{1 - \cos x}}{{\cos x\left( {1 + \cos x} \right)}}dx}=\int {\dfrac{{dm}}{m}} - \tan \left( {\dfrac{x}{2}} \right) + C$
We know that the $\int {\dfrac{{dm}}{m} = \ln m}$
$\int {\dfrac{{1 - \cos x}}{{\cos x\left( {1 + \cos x} \right)}}dx}=\int {\dfrac{{dm}}{m}} - \tan \left( {\dfrac{x}{2}} \right) + C$
Substitute the value of m,
$\therefore \int {\dfrac{{1 - \cos x}}{{\cos x\left( {1 + \cos x} \right)}}dx}= \log \left( {\sec x + \tan x} \right) - 2\tan \left( {\dfrac{x}{2}} \right) + C$
This is the final integral.

So the correct option is A.

Note: In this problem the only thing to note is the approach to solve and reach the answer. We can solve the same problem by some other methods but those might not lead towards the answer present in our options. So we need to choose a solution.