\(\int\frac{1}{\sin^2x.\cos^2x}\text{ d}x=\)
\(\int\dfrac{1}{\sin^2x.\cos^2x}\text{d}x=-2\cot2x+C\). \(\int\dfrac{1}{\sin^2x.\cos^2x}\text{d}x=-\cot2x+C\). \(\int\dfrac{1}{\sin^2x.\cos^2x}\text{d}x=-2\cot^2x+C\). \(\int\dfrac{1}{\sin^2x.\cos^2x}\text{d}x=-2\tan^2x+C\). Hướng dẫn giải:\(\int\dfrac{1}{\sin^2x.\cos^2x}\text{d}x=\int\dfrac{1}{\dfrac{1}{4}\sin^22x}\text{d}x=2\int\dfrac{1}{\sin^22x}d\left(2x\right)=-2\cot2x+C\)