Question

$ \int{\frac{{{\sin }^{2}}x}{{{\cos }^{4}}x}}dx $ is equal to:

• A. $ \frac{1}{3}{{\tan }^{2}}x+c $
• B. $ \frac{1}{2}{{\tan }^{2}}x+c $
• C. $ \frac{1}{3}{{\tan }^{3}}x+c $
• D. $ 3\sin 2x-4\cos 4x+c $
• E. $ {{\cos }^{3}}x+c $

Answer 1

Answer: (C)

Let $ I=\int{\frac{{{\sin }^{2}}x}{{{\cos }^{4}}x}}dx $ $ =\int{{{\tan }^{2}}x{{\sec }^{2}}x\,dx} $ Let $ tan\text{ }x=t $ $ \Rightarrow $ $ se{{c}^{2}}x\text{ }dx=dt $ $ \therefore $ $ I=\int{{{t}^{2}}}\,dt=\frac{{{t}^{3}}}{3}+c=\frac{{{\tan }^{3}}x}{3}+c $