Question

Evaluate the integral \[ \int_0^{\pi/2} \frac{\cos^2 x \sin^2 x}{\cos^2 x + \sin^2 x} \, dx. \]

• A. \( \frac{1}{3} \)
• B. \( \frac{1}{6} \)
• C. \( \frac{1}{4} \)
• D. \( \frac{1}{2} \)

Hint:

Using trigonometric identities like the double-angle identity can simplify integrals involving products of sine and cosine functions.

Answer 1

The Correct Option is C

Step 1: Simplifying the integrand.
We are given the integral:
\[ \int_0^{\pi/2} \frac{\cos^2 x \sin^2 x}{\cos^2 x + \sin^2 x} \, dx. \]
We know that \( \cos^2 x + \sin^2 x = 1 \) from the Pythagorean identity. Thus, the integral simplifies to:
\[ \int_0^{\pi/2} \cos^2 x \sin^2 x \, dx. \]

Step 2: Using a trigonometric identity.
We can use the double-angle identity:
\[ \sin 2x = 2 \sin x \cos x \quad \Rightarrow \quad \sin^2 2x = 4 \sin^2 x \cos^2 x. \]
Thus, the integral becomes:
\[ \int_0^{\pi/2} \frac{1}{4} \sin^2 2x \, dx. \]

Step 3: Solving the integral.
We now evaluate:
\[ \int_0^{\pi/2} \sin^2 2x \, dx. \]
Using the identity \( \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \), we get: \[ \int_0^{\pi/2} \sin^2 2x \, dx = \frac{1}{2} \int_0^{\pi/2} (1 - \cos 4x) \, dx. \]

Step 4: Evaluating the integral.
We can now solve:
\[ \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2}, \quad \int_0^{\pi/2} \cos 4x \, dx = 0 \quad \text{(since the integral of cosine over a complete period is zero)}. \]
Thus, we have:
\[ \frac{1}{2} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{4}. \]

Step 5: Final result.
Now multiply by \( \frac{1}{4} \):
\[ \frac{1}{4} \times \frac{\pi}{4} = \frac{1}{4}. \]
Final Answer:
The correct answer is: \[ \boxed{\frac{1}{4}}. \]