Question

\(\int \frac{\sec x \sqrt{\sec x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx = \)

• A. \(2(\sqrt{\tan x} - \log_e(\sqrt{\tan x} + 1)) + C\)
• B. \(2(\sqrt{\tan x} + \log_e(\sqrt{\tan x} + 1)) + C\)
• C. \(2\sqrt{\tan x} - \log_e(\sqrt{\tan x} + 1) + C\)
• D. \(2\sqrt{\tan x} + \log_e(\sqrt{\tan x} + 1) + C\)
• E. \(\sqrt{\tan x} - \log_e(\sqrt{\tan x} + 1) + C\)

Hint:

When an integrand involves roots of \(\sin x\) and \(\cos x\), dividing by a power of \(\cos x\) to convert to \(\tan x\) is a standard winning strategy.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
To solve this integration, we should express the entire integrand in terms of \(\tan x\) and its derivative \(\sec^2 x\).

Step 2: Key Formula or Approach:
Divide both numerator and denominator by \(\sqrt{\cos x}\) to introduce \(\tan x\) terms.

Step 3: Detailed Explanation:
Let \(I = \int \frac{\sec x \sqrt{\sec x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx\).
Divide numerator and denominator by \(\sqrt{\cos x}\):
\[ I = \int \frac{\frac{\sec x \sqrt{\sec x}}{\sqrt{\cos x}}}{\frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\cos x}}} dx = \int \frac{\sec x \sqrt{\sec^2 x}}{\sqrt{\tan x} + 1} dx = \int \frac{\sec^2 x}{\sqrt{\tan x} + 1} dx \]
Substitute \(u = \sqrt{\tan x}\). Then \(du = \frac{1}{2\sqrt{\tan x}} \sec^2 x dx \implies \sec^2 x dx = 2u du\).
\[ I = \int \frac{2u du}{u + 1} \]
Using long division or adjustment:
\[ I = 2 \int \frac{(u + 1) - 1}{u + 1} du = 2 \int (1 - \frac{1}{u + 1}) du \]
\[ I = 2(u - \log_e(u + 1)) + C \]
Substitute back \(u = \sqrt{\tan x}\):
\[ I = 2(\sqrt{\tan x} - \log_e(\sqrt{\tan x} + 1)) + C \]

Step 4: Final Answer:
The integral is \(2(\sqrt{\tan x} - \log_e(\sqrt{\tan x} + 1)) + C\).