Question

Find the value of $\int{x\left( \dfrac{\sec 2x-1}{\sec 2x+1} \right)dx}$?

Answer 1

For this problem we need to find the integration value of the given function. We can observe that the given function has multiplication and division operators, so we can’t directly calculate the integration value of the given function by using the known integration formulas. For this we will first convert all the trigonometric ratios into $\sin$, $\cos$ and apply some trigonometric formulas or identities and simplify the equation. Now we will get an equation which is in the form of $\int{uvdx}$, so we will apply integration by parts formula and use the integration formulas to get the required result.

Complete step-by-step solution:
Given that, $\int{x\left( \dfrac{\sec 2x-1}{\sec 2x+1} \right)dx}$.
Applying the formula $\sec x=\dfrac{1}{\cos x}$ in the above equation, then we will get
$\Rightarrow \int{x\left( \dfrac{\sec 2x-1}{\sec 2x+1} \right)dx}=\int{x\left( \dfrac{\dfrac{1}{\cos 2x}-1}{\dfrac{1}{\cos 2x}+1} \right)dx}$
Simplifying the above equation, then we will have
$\begin{aligned} & \Rightarrow \int{x\left( \dfrac{\sec 2x-1}{\sec 2x+1} \right)dx}=\int{x\left( \dfrac{\dfrac{1-\cos 2x}{\cos 2x}}{\dfrac{1+\cos 2x}{\cos 2x}} \right)dx} \\\ & \Rightarrow \int{x\left( \dfrac{\sec 2x-1}{\sec 2x+1} \right)dx}=\int{x\left( \dfrac{1-\cos 2x}{1+\cos 2x} \right)dx} \\\ \end{aligned}$
In trigonometry we have the formula $1-\cos 2x=2{{\sin }^{2}}x$, $1+\cos 2x=2{{\cos }^{2}}x$. Substituting these values in the above equation, then we will get
$\Rightarrow \int{x\left( \dfrac{\sec 2x-1}{\sec 2x+1} \right)dx}=\int{x\left( \dfrac{2{{\sin }^{2}}x}{2{{\cos }^{2}}x} \right)dx}$
We know that the value of $\dfrac{\sin x}{\cos x}=\tan x$, substituting this value in the above equation, then we will get
$\Rightarrow \int{x\left( \dfrac{\sec 2x-1}{\sec 2x+1} \right)dx}=\int{x{{\tan }^{2}}xdx}$
We have the trigonometric identity ${{\sec }^{2}}x-{{\tan }^{2}}x=1$. From the identity we can write the above equation as
$\begin{aligned} & \Rightarrow \int{x\left( \dfrac{\sec 2x-1}{\sec 2x+1} \right)dx}=\int{x\left( {{\sec }^{2}}x-1 \right)dx} \\\ & \Rightarrow \int{x\left( \dfrac{\sec 2x-1}{\sec 2x+1} \right)dx}=\int{x{{\sec }^{2}}xdx}-\int{xdx} \\\ \end{aligned}$
Applying the integration by parts rule which says that $\int{uvdx}=u\int{vdx}-\int{\left( {{u}^{'}} \right)\left( \int{vdx} \right)dx}$. We will apply the ILATE rule, which is about the order of choosing the function u. It is Inverse, Logarithm, Algebra, Trigonometry and Exponential. Here, we have trigonometric and algebraic functions. Applying this formula in the above equation, then we will get
$\Rightarrow \int{x\left( \dfrac{\sec 2x-1}{\sec 2x+1} \right)dx}=x\int{{{\sec }^{2}}xdx}-\int{{{\left( x \right)}^{'}}\left( \int{{{\sec }^{2}}xdx} \right)dx}-\int{xdx}$
Applying the known formulas $\int{{{\sec }^{2}}xdx}=\tan x+c$, $\int{xdx}=\dfrac{{{x}^{2}}}{2}+c$, $\dfrac{d}{dx}\left( x \right)=1$ in the above equation, then we will get
$\begin{aligned} & \Rightarrow \int{x\left( \dfrac{\sec 2x-1}{\sec 2x+1} \right)dx}=x\tan x-\int{1\left( \tan x \right)dx}-\dfrac{{{x}^{2}}}{2}+c \\\ & \Rightarrow \int{x\left( \dfrac{\sec 2x-1}{\sec 2x+1} \right)dx}=x\tan x-\int{\tan xdx}-\dfrac{{{x}^{2}}}{2}+c \\\ \end{aligned}$
We know that the value of $\int{\tan xdx}=\log \left| \sec x \right|+c$. Substituting this value in the above equation, then we will get
$\Rightarrow \int{x\left( \dfrac{\sec 2x-1}{\sec 2x+1} \right)dx}=x\tan x-\log \left| \sec x \right|-\dfrac{{{x}^{2}}}{2}+c$
Hence the integral value of the given function $\int{x\left( \dfrac{\sec 2x-1}{\sec 2x+1} \right)dx}$ is $x\tan x-\log \left| \sec x \right|-\dfrac{{{x}^{2}}}{2}+c$.

Note: For this problem we can follow so many ways to solve. We can also expand the given equation as $\int{\dfrac{x\sec 2x}{\sec 2x+1}dx}-\int{\dfrac{x}{\sec 2x+1}dx}$ and considering each term as individually and calculate the integral values. This method is very lengthy and we need to use a lot of formulas to simplify the equations. So, we have followed the above procedure which is short and very easy.