Question

$\int_0^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1-\sin x \cos x} d x=$

• A. $\frac{\pi}{4}$
• B. $2\pi$
• C. $0$
• D. $\frac{\pi}{2}$

Hint:

Anytime you evaluate an integral of the form $\int_0^{\frac{\pi}{2}} \frac{f(\sin x) - f(\cos x)}{g(\sin x, \cos x)} dx$ where $g(x)$ is symmetric (meaning $g(\sin x, \cos x) = g(\cos x, \sin x)$), the answer will ALWAYS be exactly $0$ due to the King's Property cancellation!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a definite integral involving trigonometric functions bounded between $0$ and $\frac{\pi}{2}$.

Step 2: Key Formula or Approach:
This problem relies on a very famous property of definite integrals, often called the "King's Rule":
$$\int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx$$ Here, $a = \frac{\pi}{2}$. By substituting $\left(\frac{\pi}{2} - x\right)$ for $x$ everywhere in the integrand, the sine functions will change to cosine, and the cosine functions will change to sine.

Step 3: Detailed Explanation:
Let the given integral be $I$:
$$I = \int_0^{\frac{\pi}{2}} \frac{\sin x - \cos x}{1 - \sin x \cos x} \, dx \quad \dots \text{(Equation 1)}$$ Apply the King's Rule substitution ($x \rightarrow \frac{\pi}{2} - x$):
$$I = \int_0^{\frac{\pi}{2}} \frac{\sin\left(\frac{\pi}{2} - x\right) - \cos\left(\frac{\pi}{2} - x\right)}{1 - \sin\left(\frac{\pi}{2} - x\right) \cos\left(\frac{\pi}{2} - x\right)} \, dx$$ Using the complementary angle identities ($\sin(\frac{\pi}{2}-x) = \cos x$ and $\cos(\frac{\pi}{2}-x) = \sin x$):
$$I = \int_0^{\frac{\pi}{2}} \frac{\cos x - \sin x}{1 - \cos x \sin x} \, dx \quad \dots \text{(Equation 2)}$$ Notice that the numerator in Equation 2 is exactly the negative of the numerator in Equation 1. We can factor out a $-1$:
$$I = -\int_0^{\frac{\pi}{2}} \frac{\sin x - \cos x}{1 - \sin x \cos x} \, dx$$ This gives us:
$$I = -I$$ Now, solve for $I$:
$$2I = 0$$ $$I = 0$$

Step 4: Final Answer:
The value of the definite integral is $0$, corresponding to option (C).