Question

$\int \frac{\sqrt{\tan x}{\sin x \cdot \cos x} \, dx = $ ______.

• A. $2\sqrt{\sec x} + c$, where c is a constant of integration
• B. $2\sqrt{\tan x} + c$, where c is a constant of integration
• C. $\frac{2}{\sqrt{\tan x}} + c$, where c is a constant of integration
• D. $\frac{2}{\sqrt{\sec x}} + c$, where c is a constant of integration

Hint:

Whenever an integral features a mix of $\sin x$ and $\cos x$ in the denominator, dividing the numerator and denominator by $\cos^2 x$ almost always magically converts the problem into a pure $\tan x$ and $\sec^2 x$ substitution setup!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
To integrate expressions involving $\tan x$, try to create a $\sec^2 x$ term in the numerator to facilitate substitution.

Step 2: Formula Application:
Divide numerator and denominator by $\cos^2 x$.

Step 3: Explanation:
$I = \int \frac{\sqrt{\tan x}}{\sin x \cos x} \cdot \frac{\sec^2 x}{\sec^2 x} \, dx = \int \frac{\sqrt{\tan x} \sec^2 x}{\frac{\sin x \cos x}{\cos^2 x}} \, dx$. $I = \int \frac{\sqrt{\tan x} \sec^2 x}{\tan x} \, dx = \int \frac{\sec^2 x}{\sqrt{\tan x}} \, dx$. Let $\tan x = t$, then $\sec^2 x \, dx = dt$. $I = \int \frac{1}{\sqrt{t}} \, dt = 2\sqrt{t} + c = 2\sqrt{\tan x} + c$.

Step 4: Final Answer:
The integral is $2\sqrt{\tan x} + c$.