Question

Evaluate the integral \[ \int \frac{\sin 2x}{\sin^2 x \cos^2 x}\, dx \]

• A. \( \log|\tan^2 x| + c \)
• B. \( \log|\sec^2 x| + c \)
• C. \( \log|\tan x| + c \)
• D. \( \log|\sec x| + c \)

Hint:

When trigonometric functions appear in powers, try simplifying using identities before integrating.

Answer 1

The Correct Option is A

Step 1: Use the identity for \( \sin 2x \).
\[ \sin 2x = 2\sin x \cos x \]

Step 2: Substitute into the integral.
\[ \int \frac{2\sin x \cos x}{\sin^2 x \cos^2 x}\, dx = \int \frac{2}{\sin x \cos x}\, dx \]

Step 3: Rewrite using tangent and secant.
\[ \frac{2}{\sin x \cos x} = 2(\tan x + \cot x) \]

Step 4: Integrate.
\[ \int 2(\tan x + \cot x)\, dx = 2(\log|\sec x| + \log|\csc x|) + c \] \[ = \log|\tan^2 x| + c \]

Step 5: Conclusion.
\[ \boxed{\log|\tan^2 x| + c} \]