Question

$\int\limits^a_b \, dx/ sin2xcos2x$

Answer 1

$\int _{a}^{b}\frac{dx}{sin2xcos2x}=\int _{a}^{b}\frac{2\, dx}{sin4x}=2\int _{a}^{b}\frac{sin4x}{sin^24x}=2\int _{a}^{b}\frac{sin4x}{1-cos^24x}=\\\\=[\, t=cos4x,dt=-4sin4x\, dx\, ]=\\\\=-\frac{1}{2}\int _{cos4a}^{cos4b}\frac{dt}{1-t^2}=\frac{1}{2}\int _{cos4a}^{cos4b}\frac{dt}{t^2-1}=\frac{1}{2}\cdot \frac{1}{2}ln|\frac{t-1}{t+1}|_{cos4a}^{cos4b}=\\\\=\frac{1}{4}(ln|\frac{cos4b-1}{cos4b+1}|-ln|\frac{cos4a-1}{cos4a+1}|)=\frac{1}{4}(ln|tg2b|-ln|tg2a|)$