Question:
$\int \frac{\sec x}{(\sec x+\tan x)^{9}}dx =$
$\int \frac{\sec x}{(\sec x+\tan x)^{9}}dx =$
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The derivative of $(\sec x + \tan x)$ is $\sec x(\sec x + \tan x)$.
Updated On: Apr 28, 2026
- $\frac{1}{9}(\sec x+\tan x)^{9}+C$
- $\frac{-1}{9}(\sec x+\tan x)^{9}+C$
- $\frac{-1}{9}(\sec x+\tan x)^{-9}+C$
- $\frac{1}{9}(\sec x+\tan x)^{-9}+C$
- $(\sec x+\tan x)^{-9}+C$
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The Correct Option is C
Solution and Explanation
Step 1: Concept
Let $u = \sec x + \tan x$.
Step 2: Analysis
$du = (\sec x \tan x + \sec^2 x) dx = \sec x (\sec x + \tan x) dx$. So, $\frac{du}{u} = \sec x dx$.
Step 3: Calculation
$\int \frac{1}{u^9} \cdot \frac{du}{u}$ is not correct here; rather, the substitution leads to a power rule application. $\int u^{-10} du = \frac{u^{-9}}{-9} + C$. Final Answer: (C)
Let $u = \sec x + \tan x$.
Step 2: Analysis
$du = (\sec x \tan x + \sec^2 x) dx = \sec x (\sec x + \tan x) dx$. So, $\frac{du}{u} = \sec x dx$.
Step 3: Calculation
$\int \frac{1}{u^9} \cdot \frac{du}{u}$ is not correct here; rather, the substitution leads to a power rule application. $\int u^{-10} du = \frac{u^{-9}}{-9} + C$. Final Answer: (C)
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