Question

Evaluate the integral: \[ \int \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx \]

• A. \(2\sqrt{\tan x} + C\)
• B. \( \frac{2}{\sin^2 x} \)
• C. \( \frac{2}{\cos x} \)
• D. \( \frac{2}{\sin x} \)

Hint:

Remember to use standard trigonometric identities to simplify complex expressions and recognize common integrals.

Answer 1

The Correct Option is A

Step 1: Write the given integral:
\[ \int \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx \]

Step 2: Express \(\tan x\) in terms of sine and cosine:
\[ \tan x = \frac{\sin x}{\cos x} \Rightarrow \sqrt{\tan x} = \frac{\sqrt{\sin x}}{\sqrt{\cos x}} \]

Step 3: Substitute into the integral:
\[ \int \frac{\frac{\sqrt{\sin x}}{\sqrt{\cos x}}}{\sin x \cos x} \, dx \]

Step 4: Simplify:
\[ = \int \frac{1}{\sin^{1/2} x \cos^{3/2} x} \, dx \]

Step 5: Use substitution:
Let \( t = \tan x \Rightarrow dt = \sec^2 x \, dx = \frac{1}{\cos^2 x} dx \)
So, \( dx = \cos^2 x \, dt \)

Step 6: Substitute everything:
\[ \int \frac{\sqrt{t}}{\sin x \cos x} \cdot \cos^2 x \, dt \]
Since \( \sin x = t \cos x \), substitute:
\[ = \int \frac{\sqrt{t} \cdot \cos^2 x}{t \cos^2 x} \, dt \]
\[ = \int \frac{1}{\sqrt{t}} \, dt \]

Step 7: Integrate:
\[ \int t^{-1/2} dt = 2\sqrt{t} + C \]

Step 8: Substitute back \( t = \tan x \):
\[ = 2\sqrt{\tan x} + C \]

Step 9: Final Answer:
\[ \boxed{2\sqrt{\tan x} + C} \]

Answer 2

Step 1: Let \( t = \sqrt{\tan x} \).
Then \( \tan x = t^2 \).

Step 2: Differentiate:
\[ \sec^2 x \, dx = 2t \, dt \Rightarrow dx = \frac{2t}{\sec^2 x} \, dt \]

Step 3: Use identity:
\[ \sec^2 x = 1 + \tan^2 x = 1 + t^4 \]

Step 4: Rewrite the integral:
\[ \int \frac{\sqrt{\tan x}}{\sin x \cos x} dx = \int \frac{t}{\sin x \cos x} \cdot \frac{2t}{1 + t^4} dt \]

Step 5: Use identity \( \sin x \cos x = \frac{\tan x}{\sec^2 x} \):
\[ \sin x \cos x = \frac{t^2}{1 + t^4} \]

Step 6: Substitute:
\[ = \int \frac{t \cdot 2t}{t^2} dt = \int 2 \, dt \]

Step 7: Integrate:
\[ = 2t + C \]

Step 8: Substitute back:
\[ = 2\sqrt{\tan x} + C \]

Final Answer: \(2\sqrt{\tan x} + C\)