lilmoore11p8

2022-07-01

Solving $\int \frac{2\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)}{{\mathrm{sin}}^{4}\left(x\right)+{\mathrm{cos}}^{4}\left(x\right)}\phantom{\rule{thinmathspace}{0ex}}dx$

Caiden Barrett

Expert

Hint
For the denominator
${\mathrm{sin}}^{4}\left(x\right)+{\mathrm{cos}}^{4}\left(x\right)=\left({\mathrm{sin}}^{2}\left(x\right)+{\mathrm{cos}}^{2}\left(x\right){\right)}^{2}-2{\mathrm{sin}}^{2}\left(x\right){\mathrm{cos}}^{2}\left(x\right)=1-\frac{1}{2}{\mathrm{sin}}^{2}\left(2x\right)=\frac{1}{2}\left(1+{\mathrm{cos}}^{2}\left(2x\right)\right)$
For the numerator
$2\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)=\mathrm{sin}\left(2x\right)$
So
$\int \frac{2\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)}{{\mathrm{sin}}^{4}\left(x\right)+{\mathrm{cos}}^{4}\left(x\right)}\phantom{\rule{thinmathspace}{0ex}}dx=\int \frac{2\mathrm{sin}\left(2x\right)}{1+{\mathrm{cos}}^{2}\left(2x\right)}\phantom{\rule{thinmathspace}{0ex}}dx$
Changing variable $t=\mathrm{cos}\left(2x\right)$ looks quite promising.

nidantasnu

Expert

$\begin{array}{rl}\int \frac{2\mathrm{sin}x\mathrm{cos}x}{{\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x& =\int \frac{2\mathrm{sin}x\mathrm{cos}x}{1-2{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\\ & =\int \frac{\mathrm{sin}2x}{1-\frac{1}{2}{\mathrm{sin}}^{2}2x}\mathrm{d}x\\ & =\int \frac{\mathrm{sin}2x}{1+{\mathrm{cos}}^{2}2x}\mathrm{d}\left(2x\right)\\ & =-\mathrm{arctan}\mathrm{cos}2x\phantom{\rule{thickmathspace}{0ex}}.\end{array}$