Question

\[\int\frac{\sin^6 x}{\cos^8 x} dx =\]

Options

•  tan 7x + C

• \[\frac{\tan^7 x}{7} + C\]
• \[\frac{\tan 7x}{7} + C\]

• sec⁷ x + C

Answer 1

\[\frac{\tan^7 x}{7} + C\]

\[\text{Let }I = \int\frac{\sin^6 x}{\cos^8 x}dx\]

\[ = \int\frac{\sin^6 x}{\cos^6 x} \times \frac{1}{\cos^2 x}dx\]

\[ = \int \tan^6 x \cdot \sec^2 x dx\]

\[\text{Putting }\tan x = t\]
\[ \Rightarrow \sec^2 x dx = dt\]
\[ \therefore I = \int t^6 \cdot dt\]
\[ = \frac{t^7}{7} + C\]
\[ = \frac{\tan^7 x}{7} + C ............\left( \because t = \tan x \right)\]