Salvatore Boone

2022-01-05

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

Barbara Meeker

Observe that, $\frac{d\left(x\mathrm{sin}x+\mathrm{cos}x\right)}{dx}=x\mathrm{cos}x$
$\int \frac{{x}^{2}dx}{{\left(x\mathrm{sin}x+\mathrm{cos}x\right)}^{2}}=\int \frac{x}{\mathrm{cos}x}\cdot \frac{x\mathrm{cos}x}{{\left(x\mathrm{sin}x+\mathrm{cos}x\right)}^{2}}dx$
So, if
So, $\int \frac{x\mathrm{cos}x}{{\left(x\mathrm{sin}x+\mathrm{sin}x\right)}^{2}}dx=\int \frac{dz}{{z}^{2}}=-\frac{1}{2}=-\frac{1}{x\mathrm{sin}x+\mathrm{cos}x}$
So,
$I=\frac{x}{\mathrm{cos}x}\int \frac{x\mathrm{cos}x}{{\left(x\mathrm{sin}x+\mathrm{cos}x\right)}^{2}}dx-\int \left(\frac{d\left(\frac{x}{\mathrm{cos}x}\right)}{dx}\int \frac{x\mathrm{cos}x}{{\left(x\mathrm{sin}x+\mathrm{cos}x\right)}^{2}}dx\right)dx$
$=-\frac{x}{\mathrm{cos}x\left(x\mathrm{sin}x+\mathrm{cos}x\right)}+\int \left(\frac{x\mathrm{sin}x+\mathrm{cos}x}{{\mathrm{cos}}^{2}x}\right)\left(\frac{1}{x\mathrm{sin}x+\mathrm{cos}x}\right)dx$
$=-\frac{x}{\mathrm{cos}x\left(x\mathrm{sin}x+\mathrm{cos}x\right)}+\int {\mathrm{sec}}^{2}xdx$
$=-\frac{x}{\mathrm{cos}x\left(x\mathrm{sin}x+\mathrm{cos}x\right\}+\mathrm{tan}x+C}$
where C is an arbitrary constant of indefinite integral
Another forn will be $\frac{\mathrm{sin}}{}$

Kayla Kline

I was inspired by this post to conduct this method.
$\int \frac{{x}^{2}}{{\left(x\mathrm{sin}x+\mathrm{cos}x\right)}^{2}}dx$
The Harmonic Addition Theorem comes in handy, so $x\mathrm{sin}x+\mathrm{cos}x=\sqrt{1+{x}^{2}}\mathrm{cos}\left(x-\alpha \right)$ where $\alpha =\mathrm{arctan}\left(x\right)$.
The origin integration becomes:
$\int \frac{{x}^{2}}{1+{x}^{2}}{\mathrm{sec}}^{2}\left(x-\alpha \right)dx$
Notice that $\int \frac{{x}^{2}}{1+{x}^{2}}dx=x-\mathrm{arctan}x=x-\alpha$
So let $t=x-\alpha$ then $dt=\frac{{x}^{2}}{1+{x}^{2}}dx$ and the integration simplifies as:
$\int {\mathrm{sec}}^{2}\left(t\right)dt=\mathrm{tan}\left(t\right)=\mathrm{tan}\left(x-\mathrm{arctan}x\right)$
In conclusion,
$\int \frac{{x}^{2}}{{\left(x\mathrm{sin}x+\mathrm{cos}x\right)}^{2}}dx=\mathrm{tan}\left(x-\mathrm{arctan}x\right)+C$

karton

Differentiation of
$x\mathrm{sin}x+\mathrm{cos}x$ is $\mathrm{cos}x$
$\int \frac{x\mathrm{cos}x}{\underset{II}{\underset{⏟}{\left(x\mathrm{sin}x+\mathrm{cos}x{\right)}^{2}}}}\cdot \frac{x}{\left(\mathrm{cos}x\right)}dx$
Now integrate by parts.
$I=\frac{-1}{\left(x\mathrm{sin}x+\mathrm{cos}x\right)}\cdot \frac{x}{\mathrm{cos}x}+\int \frac{1}{\left(x\mathrm{sin}x+\mathrm{cos}x\right)}\cdot \frac{\mathrm{cos}x.1-x\left(-\mathrm{sin}x\right)}{{\mathrm{cos}}^{2}x}$
Now, I hope things are clear to you.

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