veneciasp

2022-07-10

How to integrate $\int \frac{1}{{\mathrm{sin}}^{4}x{\mathrm{cos}}^{4}x}\phantom{\rule{thinmathspace}{0ex}}dx$?

Elianna Wilkinson

Expert

By using the substitution $x=\mathrm{arctan}t$ we have $dx=\frac{dt}{1+{t}^{2}}$ and ${\mathrm{sin}}^{2}\left(x\right)=\frac{{t}^{2}}{1+{t}^{2}}$, ${\mathrm{cos}}^{2}\left(x\right)=\frac{1}{1+{t}^{2}}$, hence
$\int \frac{dx}{{\mathrm{sin}}^{4}\left(x\right){\mathrm{cos}}^{4}\left(x\right)}=\int \frac{\left(1+{t}^{2}{\right)}^{3}}{{t}^{4}}\phantom{\rule{thinmathspace}{0ex}}dt$
and the last integral is straightforward to compute through the binomial formula. It equals:
$C-\frac{1}{3{t}^{3}}-\frac{3}{t}+3t+\frac{{t}^{3}}{3}.$

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