Question

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

• A. \(\sec x + c\)
• B. \(\log \sec x + c\)
• C. \(\frac{1}{2}(\log \sec x)^2 + c\)
• D. \(\log(\log \sec x) + c\)

Hint:

In substitution integrals, look for a composite function. The derivative of the "inner" function is often present as a factor in the integrand. Here, the inner function is \(\log(\sec x)\), and its derivative, \(\tan x\), is conveniently the other factor. Recognizing this pattern is key.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to find the indefinite integral of the function \((\log \sec x) \tan x\).

Step 2: Key Formula or Approach:
This integral can be solved using the method of substitution. We look for a function and its derivative within the integrand.
Let's test the substitution \(t = \log \sec x\). We need to find its derivative, \(\frac{dt}{dx}\).

Step 3: Detailed Explanation:
Let \(I = \int (\log \sec x) \tan x \,dx\).
Let's use the substitution \(t = \log \sec x\).
Now, we differentiate \(t\) with respect to \(x\) to find \(dt\):
\[ \frac{dt}{dx} = \frac{d}{dx}(\log \sec x) \]
Using the chain rule, let \(u = \sec x\), so \(t = \log u\).
\[ \frac{dt}{dx} = \frac{dt}{du} \cdot \frac{du}{dx} = \frac{1}{u} \cdot (\sec x \tan x) = \frac{1}{\sec x} \cdot (\sec x \tan x) = \tan x \]
So, we have \(\frac{dt}{dx} = \tan x\), which means \(dt = \tan x \,dx\).
Now we can substitute \(t\) and \(dt\) back into the integral:
\[ I = \int (\log \sec x) (\tan x \,dx) = \int t \,dt \]
This is a standard integral:
\[ \int t \,dt = \frac{t^2}{2} + c \]
Finally, substitute back \(t = \log \sec x\):
\[ I = \frac{(\log \sec x)^2}{2} + c = \frac{1}{2}(\log \sec x)^2 + c \]

Step 4: Final Answer:
The value of the integral is \(\frac{1}{2}(\log \sec x)^2 + c\).