Question

\[\int\frac{\sin^2 x}{\cos^4 x} dx =\]

Options

• \[\frac{1}{3} \tan^2 x + C\]
• \[\frac{1}{2} \tan^2 x + C\]
• \[\frac{1}{3} \tan^3 x + C\]

• none of these

Answer 1

\[\frac{1}{3} \tan^3 x + C\]

 

 

\[\text{Let }I = \int\frac{\sin^2 x dx}{\cos^4 x}\]

\[ = \int\frac{\sin^2 x}{\cos^2 x} \times \frac{1}{\cos^2 x}dx\]

\[ = \int \tan^2 x \cdot \sec^2 x dx\]

\[\text{Let }\tan x = t\]

\[ \Rightarrow \sec^2 x dx = dt\]

\[ \therefore I = \int t^2 \cdot dt\]

\[ = \frac{t^3}{3} + C\]

\[ = \frac{\tan^3 x}{3} + C ..........\left( \because t = \tan x \right)\]