Phân tích: Nhận xét: $\left( \cos x \right)'=-\sin x$ . Do đó ta có thể làm như sau:

$I=\int{\frac{{{\sin }^{2}}x.\sin x}{{{\cos }^{4}}x}dx}=-\int{\frac{1-{{\cos }^{2}}x}{{{\cos }^{4}}x}d\left( \cos x \right)}$$=-\int{\left( \frac{1}{{{\cos }^{4}}x}-\frac{1}{{{\cos }^{2}}x} \right)d\left( \cos x \right)}$

$=-\int{\left( {{\cos }^{-4}}x-{{\cos }^{-2}}x \right)d\left( \cos x \right)}$

$=-\left( \frac{1}{-4+1}.{{\cos }^{-4+1}}x-\frac{1}{1-2}.{{\cos }^{-2+1}}x \right)+C$

$=-\left( \frac{-1}{3}{{\cos }^{-3}}x+{{\cos }^{-1}}x \right)+C$

$=\frac{1}{3{{\cos }^{3}}x}-\frac{1}{\cos x}+C$