Question:medium

The value of \[ \int (\log\sec x)\tan x\,dx \] is

Show Hint

Use substitution when one factor is the derivative of another factor.
  • \(\sec x+c\)
  • \(\log\sec x+c\)
  • \(\frac{1}{2}(\log\sec x)^2+c\)
  • \(\log(\log\sec x)\)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
This integral can be solved using the method of substitution (u-substitution). The key is to identify a part of the integrand whose derivative is also present in the integrand.
Step 2: Key Formula or Approach:
We will use the substitution method. Let $u = \log(\sec x)$. Then we find $du$ in terms of $dx$ and see if it matches the rest of the integrand. We will need the derivative formula $\frac{d}{dx}(\log u) = \frac{1}{u}\frac{du}{dx}$ and $\frac{d}{dx}(\sec x) = \sec x \tan x$.
Step 3: Detailed Explanation:
The integral is $\int (\log \sec x) \tan x \,dx$. Let's choose the substitution: \[ u = \log(\sec x) \] Now, let's find the derivative of $u$ with respect to $x$: \[ \frac{du}{dx} = \frac{d}{dx}(\log(\sec x)) \] Using the chain rule: \[ \frac{du}{dx} = \frac{1}{\sec x} \cdot \frac{d}{dx}(\sec x) \] \[ \frac{du}{dx} = \frac{1}{\sec x} \cdot (\sec x \tan x) \] \[ \frac{du}{dx} = \tan x \] This means we can write $du = \tan x \,dx$. Now, substitute $u$ and $du$ back into the original integral: \[ \int (\log \sec x) \tan x \,dx = \int u \,du \] This is a simple power rule integration: \[ \int u \,du = \frac{u^2}{2} + C \] Finally, substitute back $u = \log(\sec x)$: \[ = \frac{(\log \sec x)^2}{2} + C \] Step 4: Final Answer:
The value of the integral is $\frac{1}{2}(\log \sec x)^2 + c$. Therefore, option (C) is correct.
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