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

My approach :

Dividing the denominator by $\cos^2x$ we get $\dfrac{x^2\sec^2x }{(x -\tan x)(x\tan x +1)}$ then

$$\int \frac{x^2\sec^2x}{x^2\tan x -x\tan^2x+x-\tan x}\ dx=\int \frac{x^2(1+\tan^2x)}{x^2\tan x -x\tan^2x+x-\tan x}dx$$

But I am not getting any relation between numerator and denominator so that I will take any substitution and solve further please suggest whether it is correct and how to proceed in this. Thanks.

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    $\begingroup$ Use partial fractions. $\endgroup$
    – Tunk-Fey
    Commented Jun 5, 2014 at 15:32
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    $\begingroup$ It looks like the denominator has lots of double-angle occurrences in it; have you distributed the multiplication directly? $\endgroup$
    – abiessu
    Commented Jun 5, 2014 at 15:32
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    $\begingroup$ To get trig functions in the correct font, use a backslash before them. So \sin gives $\sin$ I did the title. $\endgroup$ Commented Jun 5, 2014 at 15:42

5 Answers 5

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HINT :

Rewrite the integrand $$ \frac{x^2}{(x\cos x -\sin x)(x\sin x +\cos x)} $$ as $$ \frac{x\cos x}{x\sin x +\cos x}+\frac{x\sin x}{x\cos x -\sin x} $$ then $$ \frac{\color{red}{\sin x}+x\cos x-\color{red}{\sin x}}{x\sin x +\cos x}+\frac{\color{blue}{\cos x}+x\sin x-\color{blue}{\cos x}}{x\cos x -\sin x}. $$ Now let $u=x\sin x +\cos x$ and $v=x\cos x -\sin x$.

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  • $\begingroup$ Sir,how will we solve it further,any hints.I cannot solve it further. $\endgroup$
    – diya
    Commented Aug 13, 2015 at 6:25
  • $\begingroup$ That is cool. May I ask how you get different color font? $\endgroup$
    – Doug
    Commented Aug 23, 2022 at 20:59
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\begin{align} I &= \int\frac{x^2dx}{(x\sin x+\cos x)(x\cos x-\sin x)} \\ &= -\int\frac{x^2dx}{(1+x^2)\left(\frac{x}{\sqrt{1+x^2}}\sin x+\frac{1}{\sqrt{1+x^2}}\cos x\right)\left(\frac{1}{\sqrt{1+x^2}}\sin x-\frac{x}{\sqrt{1+x^2}}\cos x\right)}\\ &=-\int\frac{\left(1-\frac{1}{1+x^2}\right)dx}{\cos (x-\arctan x)\sin (x-\arctan x)}\\ &=-\int\frac{d(x-\arctan x)}{\cos (x-\arctan x)\sin (x-\arctan x)}\\ &=-\int\frac{d(2(x-\arctan x))}{\sin (2(x-\arctan x))}\\ &=-\ln|\tan (x-\arctan x)|+C. \end{align}

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$\bf{Another \; Solution::}$ Let $\displaystyle I = \int\frac{x^2}{(x\sin x+\cos x)\cdot (x\cos x-\sin x)}dx$

Now $\displaystyle I = \int\frac{2x^2}{(x^2\sin 2x+2x\cos^2 x-2x\sin^2 x-\sin 2x)}dx$

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

Now $\displaystyle (x^2-1)\sin 2x+2x\cos 2x = \sqrt{(x^2-1)^2+(2x)^2}\cdot \left\{\left(\frac{x^2-1}{x^2+1}\right)\sin 2x+\left(\frac{2x}{x^2+1}\right)\cos 2x\right\}$

Now We can write $\displaystyle (x^2-1)\sin 2x+2x\cos 2x = (x^2+1)\cdot \sin (2x+\alpha)$

Where $\displaystyle \displaystyle \left(\frac{x^2-1}{x^2+1}\right) = \cos \alpha$ and $\displaystyle \left(\frac{2x}{x^2+1}\right)=\sin \alpha$

and so $\displaystyle\tan \alpha = \left(\frac{2x}{x^2-1}\right)\Rightarrow \alpha = \tan^{-1}\left(\frac{2x}{x^2-1}\right)$

So Integral $\displaystyle I = \int \csc(2x+\alpha)\cdot \left(\frac{2x^2}{x^2+1}\right)dx$

Now let $\displaystyle (2x+\alpha) = t\Rightarrow \left\{2x+\tan^{-1}\left(\frac{2x}{x^2-1}\right)\right\} = t\;,$Then $\displaystyle \left(\frac{2x^2}{x^2+1}\right)dx=dt$

So Integral $\displaystyle I = \int \csc tdt = \ln \left|\tan \frac{t}{2}\right|+\mathcal{C} = \ln \left|\tan \left(x+\frac{\alpha}{2}\right)\right|+\mathcal{C} $

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$\bf{My\; Solution::\; }$Let $\displaystyle I = \int\frac{x^2}{(x\sin x+\cos x)\cdot (x\cos x-\sin x)}dx$

Let $x=\tan \theta\;,$ Then $dx = \sec^2 \theta d\theta.\;\; $ Then $$I = \displaystyle \int \frac{\tan^2 \theta\cdot \sec^2 \theta }{(\tan \theta\cdot \sin(\tan \theta)+\cos(\tan \theta) )\times (\tan \theta\cdot \cos(\tan \theta)-\sin(\tan \theta) )}d\theta$$

So $$\displaystyle I = \int\frac{\tan^2 \theta\cdot \sec^2 \theta\cdot \cos^2 \theta }{(\sin \theta\cdot \sin(\tan \theta)+\cos(\tan \theta)\cdot \cos \theta )\times (\sin \theta\cdot \cos(\tan \theta)-\sin(\tan \theta)\cdot \cos \theta )}d\theta$$

So $$\displaystyle I = \int\frac{2\tan^2 \theta}{2\cos(\theta-\tan \theta)\cdot \sin(\theta-\tan \theta)}d\theta = \int\frac{2\tan^2 \theta}{\sin (2\theta-2\tan \theta)}d\theta$$

Now Let $\displaystyle (2\theta-2\tan \theta) = u\;\;,$ Then $(2-2\sec^2 \theta)d\theta = du\Rightarrow 2\tan^2\theta d\theta = -du$ So $$\displaystyle I = -\int\frac{1}{\sin u}du = -\int \csc u du = -\ln \tan \left(\frac{u}{2}\right)+\mathbb{C}=-\ln \tan \left(\theta -\tan \theta\right)+\mathbb{C}$$

So $$\displaystyle I = \ln\left|\frac{\cos (\theta-\tan \theta)}{\sin (\theta-\tan \theta)}\right|+\mathbb{C} = \ln\left|\frac{\cos \theta \cdot \cos (\tan \theta)+\sin \theta \cdot \sin(\tan \theta)}{\sin \theta \cdot \cos (\tan \theta)-\cos \theta \cdot \sin(\tan \theta)}\right|+\mathbb{C}$$

So $$\displaystyle I = \ln\left|\frac{\sin (\tan \theta)\cdot \tan \theta+\cos(\tan \theta)}{\cos (\tan \theta)\cdot \tan \theta-\sin (\tan \theta)}\right|+\mathbb{C} = \ln \left|\frac{x\sin x+\cos x}{x\cos x-\sin x}\right|+\mathbb{C}$$

So $$\int\frac{x^2}{(x\sin x+\cos x)\cdot (x\cos x-\sin x)}dx=\ln \left|\frac{x\sin x+\cos x}{x\cos x-\sin x}\right|+\mathbb{C}$$

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$\displaystyle u=x\sin x+\cos x,v=x\cos x-\sin x \Rightarrow \displaystyle du=x\cos xdx,dv=-x\sin xdx$

$\begin{align}\displaystyle\int\frac{x^{2}dx}{(x\sin x+\cos x)(x\cos x-\sin x)}&=\int\left(\frac{x\cos x}{x\sin x+\cos x}-\frac{-x\sin x}{x\cos x-\sin x}\right)dx\\&=\displaystyle\int\frac{du}{u}-\int\frac{dv}{v}\\&=\ln\left| \frac{u}{v} \right|+c\end{align}$

$\displaystyle\int\frac{x^{2}dx}{(x\sin x+\cos x)(x\cos x-\sin x)}=\ln\left| \frac{x\sin x+\cos x}{x\cos x-\sin x} \right|+c $

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