Evaluate $\int \frac{x\tan(x) \sec(x)}{(\tan(x)-x)^2}$

Question

Find value of following integral $$\int \frac{x\tan(x) \sec(x)}{(\tan(x)-x)^2}\text{dx}$$

the numerator is $\text{d[sec(x)]}$ but that isnt work due to $x$ in denominator. First we can simplify as $$\int\frac{x\sin(x)}{(\sin(x)-x\cos(x))^2} \text{dx} = \int \frac{x}{(\sin(x)-x\cos(x))} \text{dx}+\int \frac{x^2 \cos(x)}{(\sin(x)-x\cos(x))^2} \text{dx}$$

but again its not manipulative. Suggest a useful substitution or method.

Thanks a lot!

Answer 1

Hint:

For $\displaystyle\int\frac{x\sin(x)dx}{(\sin(x)-x\cos(x))^2}$

$$\dfrac{d(\sin x-x\cos x)}{dx}=\cos x-(\cos x-x\sin x)=?$$

Answer 2

Consider the function $f(x) = \frac{1}{(\sin x - x\cos x)}$. Once differentiation gives:

$$f'(x) = \frac{-(\cos x +x\sin x-\cos x)}{(\sin x - x\cos x)^2} = \frac{-x \sin x}{(\sin x - x\cos x)^2}$$

So your integral is:

$$\int \frac{x \sin x}{(\sin x - x\cos x)^2} dx = \frac{-1}{(\sin x - x \cos x)}$$

Answer 3

\begin{array}{rcl} \displaystyle \int \dfrac{x \sec x \tan x \,dx}{(\tan x - x)^2} &=& \displaystyle\int\color{red}{\left(-x\csc x\right)}\cdot \color{blue}{\left(\dfrac{-\tan^2 x}{\left(\tan x - x\right)^2}\right)\,dx}\\ &=&\displaystyle \int \color{red}{(-x\csc x)\cdot \color{blue}{d\left(\dfrac{1}{\tan x - x}\right)}}\\ &=& \dfrac{x\csc x}{x-\tan x}-\displaystyle \int \dfrac{d(-x\csc x)}{\tan x - x}\\ &=& \dfrac{x\csc x}{x-\tan x}-\displaystyle \int \dfrac{x\csc x \cot x -\csc x}{\tan x - x}\,dx\\ &=& \dfrac{x\csc x}{x-\tan x}-\displaystyle \int \dfrac{\csc x \cot x\left(x - \tan x\right)}{\tan x - x}\,dx\\ &=& \dfrac{x\csc x}{x-\tan x} - \displaystyle \int \left(-\csc x \cot x\right)\,dx\\ &=& \dfrac{x\csc x}{x -\tan x} - \csc x + C\\ &=& \csc x \left(\dfrac{x}{x - \tan x} - 1\right)+C\\ &=& \dfrac{\csc x\tan x}{x - \tan x} + C\\ &=& \dfrac{\sec x}{x - \tan x} + C \end{array}