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 Question:
We must integrate a trigonometric function involving a square root of tangent. The standard approach is to convert the denominator to match the derivative of tangent ($\sec^2 x$).

Step 2: Detailed Explanation:
Let the integral be:
$I = \int \frac{\sqrt{\tan x}}{\sin x \cos x} dx$
To create a $\sec^2 x$ term (which is the derivative of $\tan x$), divide both the numerator and the denominator by $\cos^2 x$:
$I = \int \frac{\frac{\sqrt{\tan x}}{\cos^2 x}}{\frac{\sin x \cos x}{\cos^2 x}} dx$
Simplify the denominator:
$\frac{\sin x \cos x}{\cos^2 x} = \frac{\sin x}{\cos x} = \tan x$
Simplify the numerator:
$\frac{1}{\cos^2 x} = \sec^2 x$
Substitute these back into the integral:
$I = \int \frac{\sqrt{\tan x} \cdot \sec^2 x}{\tan x} dx$
Simplify the tangent terms:
$\frac{\sqrt{\tan x}}{\tan x} = \frac{1}{\sqrt{\tan x}}$
So, the integral becomes:
$I = \int \frac{\sec^2 x}{\sqrt{\tan x}} dx$
Now perform $u$-substitution:
Let $u = \tan x$
Then $du = \sec^2 x dx$
Substitute into the integral:
$I = \int \frac{1}{\sqrt{u}} du$
$I = \int u^{-1/2} du$
Integrate using the power rule:
$I = \frac{u^{1/2}}{1/2} + c$
$I = 2\sqrt{u} + c$
Substitute back $u = \tan x$:
$I = 2\sqrt{\tan x} + c$

Step 3: Final Answer:
The integral evaluates to $2\sqrt{\tan x} + c$, matching option (b).