Question

$\int \frac{\sec x}{(\sec x+\tan x)^{9}}dx =$

• A. $\frac{1}{9}(\sec x+\tan x)^{9}+C$
• B. $\frac{-1}{9}(\sec x+\tan x)^{9}+C$
• C. $\frac{-1}{9}(\sec x+\tan x)^{-9}+C$
• D. $\frac{1}{9}(\sec x+\tan x)^{-9}+C$
• E. $(\sec x+\tan x)^{-9}+C$

Hint:

The derivative of $(\sec x + \tan x)$ is $\sec x(\sec x + \tan x)$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We are asked to integrate a trigonometric function. This integral is best solved using the method of substitution, by identifying a function and its derivative within the integrand.
Step 2: Key Formula or Approach:
Let's try the substitution \( u = \sec x + \tan x \). First, find the derivative of u: \[ \frac{du}{dx} = \frac{d}{dx}(\sec x + \tan x) = \sec x \tan x + \sec^2 x \] Factor out \( \sec x \): \[ \frac{du}{dx} = \sec x (\tan x + \sec x) = \sec x \cdot u \] From this, we can write \( du = u \sec x dx \), which gives us an expression for \( \sec x dx \): \[ \sec x dx = \frac{du}{u} \] Step 3: Detailed Explanation:
The integral is \( \int \frac{\sec x dx}{(\sec x + \tan x)^8} \).
Substitute \( u = \sec x + \tan x \) and \( \sec x dx = \frac{du}{u} \):
\[ \int \frac{1}{u^8} \cdot \frac{du}{u} = \int \frac{1}{u^9} du = \int u^{-9} du \] Now, use the power rule for integration \( \int u^n du = \frac{u^{n+1}}{n+1} + C \):
\[ \int u^{-9} du = \frac{u^{-9+1}}{-9+1} + C = \frac{u^{-8}}{-8} + C = -\frac{1}{8}u^{-8} + C \] Substitute back \( u = \sec x + \tan x \):
\[ -\frac{1}{8}(\sec x + \tan x)^{-8} + C \] Analysis of Discrepancy: The calculated answer is \( -\frac{1}{8}(\sec x + \tan x)^{-8} + C \). None of the options match this result. The provided correct answer is (C) \( -\frac{1}{9}(\sec x + \tan x)^{-9} + C \). This would be the result if the original integral was \( \int \frac{\sec x}{(\sec x + \tan x)^{10}} dx \), because that would lead to \( \int u^{-11} du \), which is incorrect as well. Let's check the derivative of option (C): Let \( y = -\frac{1}{9}(\sec x + \tan x)^{-9} \). Let \( u = \sec x + \tan x \), so \( u' = u \sec x \). \( \frac{dy}{dx} = -\frac{1}{9} \cdot (-9)u^{-10} \cdot u' = u^{-10} \cdot (u \sec x) = u^{-9} \sec x = \frac{\sec x}{(\sec x + \tan x)^9} \). This means the integral in the question should have had a power of 9 in the denominator to yield answer (C). There is an error in the question or the answer key. Step 4: Final Answer:
The correct integral of the given function is \( -\frac{1}{8}(\sec x + \tan x)^{-8} + C \). The provided options and answer key appear to be incorrect.