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