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: Look at what changes inside the integral.
Call the integral $I$. The denominator $1-\sin x \cos x$ stays the same if we swap $\sin x$ and $\cos x$. The numerator $\sin x - \cos x$ just flips its sign on that swap. That hint tells us to use the reflection trick.

Step 2: Replace $x$ with $\tfrac{\pi}{2}-x$.
For limits $0$ to $\tfrac{\pi}{2}$ we may write $\int_0^{a} f(x)\,dx = \int_0^{a} f(a-x)\,dx$. Here $\sin x$ becomes $\cos x$ and $\cos x$ becomes $\sin x$. So \[ I = \int_0^{\frac{\pi}{2}} \frac{\cos x-\sin x}{1-\sin x \cos x}\,dx. \]

Step 3: Add the two forms.
The new top is exactly the negative of the old top, so this is $-I$. We now have $I = -I$.

Step 4: Solve.
Adding gives $2I = 0$, so $I = 0$. \[ \boxed{0} \]