Question

The value of the integral $\int{\dfrac{\cos 2x-1}{\cos 2x+1}dx}$ is equal to?
(a) $\tan x-x+c$
(b) $x+\tan x+c$
(c) $x-\tan x+c$
(d) $-x-\tan x+c$

Answer 1

First of all simplify the function inside the integral. Use the half angle trigonometric identities $1-\cos 2x=2{{\sin }^{2}}x$ and $1+\cos 2x=2{{\cos }^{2}}x$ to simplify the numerator and denominator respectively. Now, use the trigonometric identity ${{\sec }^{2}}x=1+{{\tan }^{2}}x$ and break the integral into two parts. Finally, use the basic formulas $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$ and $\int{{{\sec }^{2}}xdx}=\tan x$ to get the answer. Add the constant of integration ‘c’ in the end to choose the correct option.

Complete step by step answer:
Here we are asked to find the integral of the function$\dfrac{\cos 2x-1}{\cos 2x+1}$. First let us simplify the trigonometric function using the half angle formulas. Let us assume the integral as I, so we have,
$\Rightarrow I=\int{\dfrac{\cos 2x-1}{\cos 2x+1}dx}$
Using the half angle trigonometric identities given as $1-\cos 2x=2{{\sin }^{2}}x$ and $1+\cos 2x=2{{\cos }^{2}}x$ to simplify the numerator and denominator respectively, we get,
$\begin{aligned} & \Rightarrow I=\int{\dfrac{-2{{\sin }^{2}}x}{2{{\cos }^{2}}x}dx} \\\ & \Rightarrow I=-\int{\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}dx} \\\ \end{aligned}$
Using the conversion $\dfrac{\sin x}{\cos x}=\tan x$ we get,
$\Rightarrow I=-\int{{{\tan }^{2}}dx}$
Using the trigonometric identity ${{\sec }^{2}}x=1+{{\tan }^{2}}x$ and breaking the integral into parts we get,

& \Rightarrow I=-\int{\left( {{\sec }^{2}}x-1 \right)dx} \\\ & \Rightarrow I=-\left[ \int{{{\sec }^{2}}xdx}-\int{1dx} \right] \\\ \end{aligned}$$ Now, we can write the constant 1 as ${{x}^{0}}$ so we get, $$\Rightarrow I=-\left[ \int{{{\sec }^{2}}xdx}-\int{{{x}^{0}}dx} \right]$$ So applying the formulas $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$, $\int{{{\sec }^{2}}xdx}=\tan x$ we get, $$\begin{aligned} & \Rightarrow I=-\left[ \tan x-\dfrac{{{x}^{0+1}}}{0+1} \right] \\\ & \Rightarrow I=-\left[ \tan x-x \right] \\\ & \therefore I=x-\tan x+c \\\ \end{aligned}$$ Here ‘c’ is the constant of integration as we are evaluating an indefinite integral. **So, the correct answer is “Option c”.** **Note:** Note that we don’t have any direct formula for the integration of the function ${{\tan }^{2}}x$ and that is why we have converted it into the secant function. Remember the half angle trigonometric formulas and they are very helpful in solving problems of integrals with the help of substitution. Note that you cannot apply the formula $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$ for n = -1 because in that case we have the function $\dfrac{1}{x}$ whose integral is $\ln x$. Remember the formulas of integral and differential of all the trigonometric functions.