Question

Evaluate the given integral.
$\int{\dfrac{\cos 2x-1}{\cos 2x+1}dx}$
a. $\tan x-x+C$
b. $x+\tan x+C$
c. $x-\tan x+C$
d. $-x-\cot x+C$

Answer 1

Hint: We can see all the answers most of them have $\tan x$. In order to convert the given function we need to convert the numerator to $\sin x$ form and the denominator to $\cos x$ form.

Complete step-by-step answer:

In order to convert the given function in the form of $\tan x$ we need a numerator in $\sin x$ form.

We can see we have a $\cos 2x$ in numerator which can be converted into ${{\sin }^{2}}x$ form using the formula,
\[\begin{align}
  & \cos 2x=1-2{{\sin }^{2}}x \\
 & \Rightarrow \cos 2x-1=-2{{\sin }^{2}}x........(i) \\
\end{align}\]

Similarly, we can see we have a $\cos 2x$ in numerator which can be converted into ${{\cos }^{2}}x$ form using the formula,
\[\begin{align}
  & \cos 2x=2{{\cos }^{2}}x-1 \\
 & \Rightarrow \cos 2x+1=2{{\cos }^{2}}x........(ii) \\
\end{align}\]

Now, substituting the values from equation (ii) and (iii) in the given integral, we can write,
$\int{\dfrac{\cos 2x-1}{\cos 2x+1}dx}=\int{\dfrac{-2{{\sin }^{2}}x}{2{{\cos }^{2}}x}dx}$
\[\Rightarrow \int{\dfrac{\cos 2x-1}{\cos 2x+1}dx}=\int{\dfrac{-{{\sin }^{2}}x}{{{\cos }^{2}}x}dx}\]
\[\Rightarrow \int{\dfrac{\cos 2x-1}{\cos 2x+1}dx}=\int{-{{\left( \dfrac{\sin x}{\cos x} \right)}^{2}}dx}.........(iii)\]

Now, as we know, $\tan x=\left( \dfrac{\sin x}{\cos x} \right)$
Substituting this in equation (iii) we get,
\[\Rightarrow \int{\dfrac{\cos 2x-1}{\cos 2x+1}dx}=\int{-{{\tan }^{2}}xdx}........(iv)\]

Again, we know from the trigonometric identity that, ${{\tan }^{2}}x={{\sec }^{2}}x-1$
Substituting in equation (iv) we get,
\[\begin{align}
  & \Rightarrow \int{\dfrac{\cos 2x-1}{\cos 2x+1}dx}=\int{-\left( {{\sec }^{2}}x-1 \right)dx} \\
 & \Rightarrow \int{\dfrac{\cos 2x-1}{\cos 2x+1}dx}=\int{\left( 1-{{\sec }^{2}}x \right)dx} \\
 & \Rightarrow \int{\dfrac{\cos 2x-1}{\cos 2x+1}dx}=\int{dx}-\int{{{\sec }^{2}}xdx}.............(v) \\
\end{align}\]

Now we know that $\int{{{\sec }^{2}}xdx=\tan x+C}$

Substituting the values in equation (v), we get,
\[\Rightarrow \int{\dfrac{\cos 2x-1}{\cos 2x+1}dx}=x-\tan x+C\]

The integral $\int{\dfrac{\cos 2x-1}{\cos 2x+1}dx}$ leads to $x-\tan x+C$ after evaluation.

The final answer is $x-\tan x+C$.

Hence, the correct answer is option (c).

Note: The caution must be taken when substituting the values of $\cos 2x$ . Because it has various forms like $2{{\cos }^{2}}x-1,1-2{{\sin }^{2}}x,{{\cos }^{2}}x-{{\sin }^{2}}x$ and when substituting it must be assured that the correct form is substituted according to the situation. Moreover, the use of trigonometric ratios and trigonometric identities must be substituted with accordance to the requirement only. The chances of doing mistake is if \[\int{{{\sec }^{2}}xdx={{\tan }^{2}}x}\] is integrated in this form which is actually wrong.