Question

The Value of \[\int {\dfrac{{(\cos 2x - 1)}}{{(\cos 2x + 1)}}.} dx\] is
A.$\tan x - x + c$
B.$\tan x + x + c$
C.$x - \tan x + c$
D.$ - x - \cot x + c$

Answer 1

Hint: Before applying integration, we simplify the integrand by using the trigonometric identities. To apply the trigonometric identity, we take -1 common in the numerator of the integrand. In this particular problem, we use the Pythagoras trigonometric identities that is \[{\sin ^2}x + {\cos ^2}x = 1\], \[{\tan ^2}x + 1 = {\sec ^2}x\] and \[{\cot ^2}x + 1 = {\csc ^2}x\].

Complete step-by-step answer:
Given \[\int {\dfrac{{(\cos 2x - 1)}}{{(\cos 2x + 1)}}.} dx\].
In $\left( {1 - \cos 2x} \right)$, we take -1 common we have:
\[\int {\dfrac{{(\cos 2x - 1)}}{{(\cos 2x + 1)}}.} dx = - 1\int {\dfrac{{(1 - \cos 2x)}}{{(\cos 2x + 1)}}.} dx\]
\[ = - 1\int {\dfrac{{(1 - \cos 2x)}}{{(1 + \cos 2x)}}.} dx\]
Now we know trigonometric identity ${\sin ^2}x = \dfrac{{1 - \cos (2x)}}{2}$ and ${\cos ^2}x = \dfrac{{1 + \cos (2x)}}{2}$.
Using these identities we have $2{\sin ^2}x = 1 - \cos (2x)$ and $2{\cos ^2}x = 1 + \cos (2x)$. Now substituting this in above integral we have
\[ = - 1\int {\dfrac{{2{{\sin }^2}x}}{{2{{\cos }^2}x}}.} dx\]
Now cancelling 2 we have
\[ = - 1\int {\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}.} dx\]
We know that the tangent is the ratio of sine function to the cosine function. Hence we can write $\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} = {\tan ^2}x$, then we have
\[ = - 1\int {{{\tan }^2}x.} dx\]
But we don’t know the direct integration formula of ${\tan ^2}x$. So we use the identity \[{\tan ^2}x = {\sec ^2}x - 1\].
\[ = - 1\int {({{\sec }^2}x - 1).} dx\]
Now splitting the integral
\[ = - 1\int {({{\sec }^2}x).} dx - 1\int { - 1{\text{ }}dx} \]
We know the product of two negative numbers results in a positive number.
\[ = - 1\int {{{\sec }^2}x{\text{ }}} dx + 1\int {1{\text{ }}dx} \]
Now we know the integration of \[{\sec ^2}x\] is \[\tan x\], then we have
\[ = - \tan x + x + c\]
Or
\[\int {\dfrac{{(\cos 2x - 1)}}{{(\cos 2x + 1)}}.} dx = x - \tan x + c\], Where ‘c’ is the integration constant.
Hence the correct option is (c).
So, the correct answer is “Option C”.

Note: We need to simplify the integrand before applying the integration. In above after the simplification if we have \[{\cot ^2}x\] instead of \[{\tan ^2}x\], to solve this we use the Pythagoras trigonometric identity \[{\cot ^2}x + 1 = {\csc ^2}x\]. Because we know the integration of \[ - {\csc ^2}x\] is $\cot x$. Also, the integration of constant is not zero, we should be careful regarding this.