Question

Evaluate the integral \[ \int \frac{\sin^6 x}{\cos^8 x} \, dx. \]

• A. \( \tan 7x + c \)
• B. \( \frac{\tan^7 x}{7} + c \)
• C. \( \frac{\tan 7x}{7} + c \)
• D. \( \sec^7 x \)

Hint:

For integrals of the form \( \frac{\sin^m x}{\cos^n x} \), express in terms of \( \tan x \) and use substitution.

Answer 1

The Correct Option is B

Step 1: Substituting in terms of tan Rewriting in terms of \( \tan x \): \[ I = \int \frac{\sin^6 x}{\cos^8 x} dx. \] Using \( \sin x = \tan x \cos x \), we get: \[ I = \int \tan^6 x \sec^2 x dx. \] Step 2: Using substitution Let \( u = \tan x \), then \( du = \sec^2 x dx \). The integral simplifies to: \[ I = \int u^6 du. \] Step 3: Evaluating the integral \[ I = \frac{u^7}{7} + c = \frac{\tan^7 x}{7} + c. \]

Answer 2

Evaluate the integral:
\[ I = \int \frac{\sin^6 x}{\cos^8 x} \, dx \]

Step 1: Rewrite the integrand in terms of tangent and secant:
\[ \frac{\sin^6 x}{\cos^8 x} = \frac{\sin^6 x}{\cos^6 x} \cdot \frac{1}{\cos^2 x} = \tan^6 x \sec^2 x \]

Step 2: Substitute:
\[ t = \tan x \implies dt = \sec^2 x \, dx \]

Step 3: Rewrite integral in terms of \( t \):
\[ I = \int t^6 \, dt = \frac{t^7}{7} + C = \frac{\tan^7 x}{7} + C \]

Therefore, the integral evaluates to:
\[ \boxed{\frac{\tan^7 x}{7} + C} \]