Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Indefinite Integration

Question:

\(\int \frac{\sin^6x}{\cos^8x}dx=\)

Options:

\(\frac{\tan^6x}{6}+c\)

\(\frac{\tan^7x}{7}+c\)

\(\frac{\tan^5x}{5}+c\)

\(\frac{\tan^2x}{2}+c\)

Correct Answer:

\(\frac{\tan^7x}{7}+c\)

Explanation:

\(\int\frac{\sin^6x}{\cos^8x}dx=\int \tan^6x\sec^2xdx\)

Put \(\tan x=t\)