Q.By using the properties of definite integrals, evaluate the integral $\int_{0}^{\pi/2}\frac{\sin x-\cos x}{1+\sin x\cos x}\,dx$

Concept: Definite Integral Symmetry — use the substitution $x \to \frac{\pi}{2} - x$ to exploit symmetry about the midpoint.

Let $I = \int_{0}^{\pi/2} \frac{\sin x - \cos x}{1 + \sin x \cos x} \, dx$.

Substitute $x = \frac{\pi}{2} - t$. Then $\sin x = \cos t$, $\cos x = \sin t$, $dx = -dt$, and the limits swap: …

The key insight here is symmetry — but not the usual symmetry about $x = 0$. The interval is $[0, \pi/2]$, and the integrand involves both $\sin x$ and $\cos x$, which swap roles when $x$ is replaced by $\pi/2 - x$. That substitution is the classic trick for integrals over $[0, \pi/2]$ with mixed sine and cosine terms.

Let’s see why this works.

1. Set up the substitution. Let $I = \int_{0}^{\pi/2} \frac{\sin x - \cos x}{1 + \sin x \cos x} \, dx$. Use the substitution $x \to \pi/2 - x$. Then $dx$ becomes $-dx$, and the limits swap: when $x = 0$, the new variable is $\pi/2$; when $x = \pi/2$, it’s $0$. So:

2. Notice the relationship. The denominator is unchanged because $\sin(\pi/2 - x)\cos(\pi/2 - x) = \cos x \sin x = \sin x \cos x$. The numerator becomes $\cos x - \sin x = -(\sin x - \cos x)$. So we have:

3. Solve the equation. From $I = -I$, we get $2I = 0$, so $I = 0$. …

Method: Reflection giving $I=-I$ — forced-zero integrals

If reflecting the integrand reproduces its negative, then $I=-I$, which forces $I=0$ — a fast route for integrands anti-symmetric about the interval's midpoint.

Steps

Step 1: Reflect $x\to a-x$ (here $\tfrac{\pi}{2}-x$).

Apply $I=\displaystyle\int_0^a f(a-x)\,dx$ and use co-function identities ($\sin\leftrightarrow\cos$).

Step 2: Check whether the reflected integrand is $-f(x)$. …

Common Mistakes

Mistake 1: Reaching for the $t=\tan(x/2)$ Weierstrass substitution.

Why it's wrong: it produces a messy rational integral, when reflection $x\to\tfrac{\pi}{2}-x$ shows the integrand equals its own negative and the answer is $0$. Correct approach: reflect and observe $I=-I$.

Mistake 2: Not noticing the denominator is reflection-symmetric. …