piarepm

2022-01-05

Integrate:
$\int \frac{{x}^{2}}{\left(x\mathrm{cos}x-\mathrm{sin}x\right)\left(x\mathrm{sin}x+\mathrm{cos}x\right)}dx$

Kayla Kline

My solution:
Let $I=\int \frac{{x}^{2}}{\left(x\mathrm{sin}x+\mathrm{cos}x\right)\cdot \left(x\mathrm{cos}x-\mathrm{sin}x\right)}dx$
So
$I=\int \frac{{\mathrm{tan}}^{2}\theta \cdot {\mathrm{sec}}^{2}\theta }{\left(\mathrm{tan}\theta \cdot \mathrm{sin}\left(\mathrm{tan}\theta \right)+\mathrm{cos}\left(\mathrm{tan}\theta \right)\right)×\left(\mathrm{tan}\theta \cdot \mathrm{cos}\left(\mathrm{tan}\theta \right)-\mathrm{sin}\left(\mathrm{tan}\theta \right)\right)}d\theta$
So
$I=\int \frac{{\mathrm{tan}}^{2}\theta \cdot {\mathrm{sec}}^{2}\theta {\mathrm{cos}}^{2}\theta }{\left(\mathrm{sin}\theta \cdot \mathrm{sin}\left(\mathrm{tan}\theta \right)+\mathrm{cos}\left(\mathrm{tan}\theta \right)\cdot \mathrm{cos}\theta \right)}$
So
$I=\int \frac{2{\mathrm{tan}}^{2}\theta }{2\mathrm{cos}\left(\theta -\mathrm{tan}\theta \right)\cdot \mathrm{sin}\left(\theta -\mathrm{tan}\theta \right)}d\theta =\int \frac{2{\mathrm{tan}}^{2}\theta }{\mathrm{sin}\left(2\theta -2\mathrm{tan}\theta \right)}d\theta$
Now Let $\left(2\theta -2\mathrm{tan}\theta \right)=u$, then $\left(2-2{\mathrm{sec}}^{2}\theta \right)d\theta =du⇒2{\mathrm{tan}}^{2}\theta d\theta =-du$
So
$I=-\int \frac{1}{\mathrm{sin}u}du=-\int \mathrm{csc}udu$
$=-\mathrm{ln}\mathrm{tan}\left(\frac{u}{2}\right)+\mathbb{C}$
$=-\mathrm{ln}\mathrm{tan}\left(\theta -\mathrm{tan}\theta \right)+\mathbb{C}$
So
$I=\mathrm{ln}|\frac{\mathrm{cos}\left(\theta -\mathrm{tan}\theta \right)}{\mathrm{sin}\left(\theta -\mathrm{tan}\theta \right)}|+\mathbb{C}$
$=\mathrm{ln}|\frac{\mathrm{cos}\theta \cdot \left(\mathrm{tan}\theta \right)+\mathrm{sin}\theta \cdot \mathrm{sin}\left(\mathrm{tan}\theta \right)}{\mathrm{sin}\theta \cdot \mathrm{sin}\left(\mathrm{tan}\theta \right)}|+C$
So

Heather Fulton

Another Solution:
Let $I=\int \frac{{x}^{2}}{\left(x\mathrm{sin}x+\mathrm{cos}x\right)\cdot \left(x\mathrm{cos}x-\mathrm{sin}x\right)}dx$
So $I=\int \frac{2{x}^{2}}{\left({x}^{2}-1\right)\mathrm{sin}2x+2x\mathrm{cos}2x}dx$
Now $\left({x}^{2}-1\right)\mathrm{sin}2x+2x\mathrm{cos}2x=\sqrt{{\left({x}^{2}-1\right)}^{2}+{\left(2x\right)}^{2}}\cdot \left[\left(\frac{{x}^{2}-1}{{x}^{2}-1}\right)\mathrm{sin}2x+\left(\frac{2x}{{x}^{2}+1}\right)\mathrm{cos}2x\right]$
Now we can write $\left({x}^{2}-1\right)\mathrm{sin}2x+2x\mathrm{cos}2x=\left({x}^{2}+1\right)\cdot \mathrm{sin}\left(2x+\alpha \right)$
Where $\left(\frac{{x}^{2}-1}{{x}^{2}+1}\right)=\mathrm{cos}\alpha$ and $\left(\frac{2x}{{x}^{2}+1}\right)=\mathrm{sin}\alpha$
and so $\mathrm{tan}\alpha =\left(\frac{2x}{{x}^{2}-1}\right)⇒\alpha ={\mathrm{tan}}^{-1}\left(\frac{2x}{{x}^{2}-1}\right)$
So Integral $I=\int \mathrm{csc}\left(2x+\alpha \right)\cdot \left(\frac{2{x}^{2}}{{x}^{2}+1}\right)dx$
Now let $\left(2x+\alpha \right)=t⇒\left[2x+{\mathrm{tan}}^{-1}\left(\frac{2x}{{x}^{2}-1}\right)\right]=t$, then $\left(\frac{2{x}^{2}}{{x}^{2}+1}\right)dx=dt$
So integral $I=\int \mathrm{csc}tdt=\mathrm{ln}|\mathrm{tan}\frac{t}{2}|+C=\mathrm{ln}|\mathrm{tan}\left(x+\frac{\alpha }{2}\right)|+C$

star233

HINT :
Rewrite the integrand
$\frac{{x}^{2}}{\left(x\mathrm{cos}x-\mathrm{sin}x\right)\left(x\mathrm{sin}x+\mathrm{cos}x\right)}$
as
$\frac{x\mathrm{cos}x}{x\mathrm{sin}x+\mathrm{cos}x}+\frac{x\mathrm{sin}x}{x\mathrm{cos}x-\mathrm{sin}x}$
then
$\frac{\mathrm{sin}x+x\mathrm{cos}x-\mathrm{sin}x}{x\mathrm{sin}x+\mathrm{cos}x}+\frac{\mathrm{cos}x+x\mathrm{sin}x-\mathrm{cos}x}{x\mathrm{cos}x-\mathrm{sin}x}$
Now let $u=x\mathrm{sin}x+\mathrm{cos}x$ and $v=x\mathrm{cos}x-\mathrm{sin}x$

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