Q.Evaluate the definite integral: $\int_{0}^{\pi} \left(\sin^2 \frac{x}{2} - \cos^2 \frac{x}{2}\right) \, dx$

The key idea is to simplify the integrand using a trigonometric identity before integrating.

First, recall the double-angle identity: $\cos^2 \theta - \sin^2 \theta = \cos 2\theta$. Here we have $\sin^2 \frac{x}{2} - \cos^2 \frac{x}{2} = -(\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}) = -\cos x$.

So the integral becomes:

$\int_{0}^{\pi} (-\cos x) \, dx = -\int_{0}^{\pi} \cos x \, dx.$ …

We start with the integral:

The expression inside looks like it could be a cosine of a double angle — but with a sign reversed. Recall the identity:

Here we have $\sin^2 \frac{x}{2} - \cos^2 \frac{x}{2}$, which is exactly $-(\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}) = -\cos x$.

So the integrand simplifies dramatically:

Now the integral becomes:

Let’s evaluate step by step.

1. Antiderivative of $\cos x$ is $\sin x$. So:

2. Evaluate at the limits:

So:

Thus the value of the integral is $0$. …

Method: Simplify with a trigonometric identity before integrating

If an integrand is a combination of trig squares or products, first collapse it to a single simpler trig function using an identity; the integration that follows is then routine.

Steps

Step 1: Recognise the identity behind the expression.

Match the integrand to a double-angle or Pythagorean identity — for instance $\cos^2\theta-\sin^2\theta=\cos 2\theta$, so $\sin^2\theta-\cos^2\theta=-\cos 2\theta$.

Step 2: Rewrite the integrand in its reduced single-function form. …

Common Mistakes

Mistake 1: Getting the sign wrong on the identity.

Why it's wrong: $\cos^2\theta-\sin^2\theta=\cos 2\theta$, so $\sin^2\theta-\cos^2\theta=-\cos 2\theta$; here with $\theta=\tfrac x2$ the integrand is $-\cos x$, not $+\cos x$. Correct approach: factor out the minus sign explicitly before integrating.

Mistake 2: Assuming the answer must be nonzero because there is "area". …