Question

The indefinite integral $\int \frac{\sec^8 x}{\csc x}\,dx$ is equal to

• A. $\frac{\sec^8 x}{8} + c$
• B. $\frac{\sec^6 x}{6} + c$
• C. $\frac{\sec^7 x}{7} + c$
• D. $\frac{\sec^9 x}{9} + c$

Hint:

Alternatively, you can substitute $t = \cos x$ directly into $\int \frac{\sin x}{\cos^8 x} \, dx$. Since $dt = -\sin x \, dx$, the expression transforms into: $$\int -t^{-8} \, dt = -\frac{t^{-7}}{-7} + c = \frac{1}{7\cos^7 x} + c = \frac{\sec^7 x}{7} + c$$ Both methods are fast, so pick whichever substitution path your mind spots first!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem requires finding the antiderivative of a trigonometric fraction containing high powers of secant and a cosecant function.

Step 2: Key Formula or Approach:
1. Express all trigonometric functions in terms of basic sine and cosine components to spot potential substitutions. 2. Look for a function-derivative pair to apply $u$-substitution, specifically identifying the derivative formula $d(\sec x) = \sec x \tan x \, dx$.

Step 3: Detailed Explanation:
Let's rewrite the integrand using basic identity relationships ($\sec x = \frac{1}{\cos x}$ and $\csc x = \frac{1}{\sin x}$): $$I = \int \frac{\left(\frac{1}{\cos^8 x}\right)}{\left(\frac{1}{\sin x}\right)} \, dx = \int \frac{\sin x}{\cos^8 x} \, dx$$ Let's break apart the denominator terms to isolate a recognizable tangent and secant product: $$I = \int \frac{1}{\cos^7 x} \cdot \frac{\sin x}{\cos x} \, dx = \int \sec^7 x \tan x \, dx$$ To configure this perfectly for substitution, factor out one secant term: $$I = \int \sec^6 x \cdot (\sec x \tan x) \, dx$$ Now, substitute $u = \sec x$, which means $du = \sec x \tan x \, dx$: $$I = \int u^6 \, du$$ Integrating using the power rule: $$I = \frac{u^7}{7} + c$$ Substituting $u = \sec x$ back gives: $$I = \frac{\sec^7 x}{7} + c$$

Step 4: Final Answer:
The integral evaluates cleanly to $\frac{\sec^7 x}{7} + c$, which corresponds to option (C).