Q.Evaluate the definite integral: $\int_{0}^{\pi} \cos^2 x \, dx$

The key idea is to use the symmetry of $\cos^2 x$ about $\pi/2$, or equivalently, the identity $\cos^2 x = \frac{1+\cos 2x}{2}$.

Step 1: Use the double-angle identity to rewrite the integrand:

Step 2: Substitute into the integral and split:

$\int_{0}^{\pi} \cos^2 x \, dx = \frac{1}{2} \int_{0}^{\pi} 1 \, dx + \frac{1}{2} \int_{0}^{\pi} \cos 2x \, dx.$ …

The key to solving $\int_0^\pi \cos^2 x \, dx$ efficiently is noticing a symmetry — not just any symmetry, but the fact that $\cos^2 x$ is symmetric about the midpoint $x = \pi/2$. This is a classic trick for definite integrals of even powers of sine and cosine over a full period or half-period.

Why does this matter? Because instead of integrating over the whole interval $[0, \pi]$, we can double the integral over $[0, \pi/2]$, where $\cos^2 x$ has a well-known antiderivative. Let’s walk through it.

1. Recognize the symmetry. The function $\cos^2 x$ is periodic with period $\pi$, but more importantly, it is symmetric about $x = \pi/2$ on $[0, \pi]$. That is, $\cos^2(\pi - x) = \cos^2 x$. This means the area under the curve from $0$ to $\pi/2$ is exactly the same as from $\pi/2$ to $\pi$. So:

2. Use the double-angle identity. To integrate $\cos^2 x$, we rewrite it using $\cos^2 x = \frac{1 + \cos 2x}{2}$. This is the standard trick — it turns a squared trig function into something linear in $\cos 2x$, which is easy to integrate.

3. Integrate term by term.

The first integral is straightforward: $\int_0^{\pi/2} 1 \, dx = \pi/2$.

For the second, let $u = 2x$, so $du = 2 dx$, and when $x = 0$, $u = 0$; when $x = \pi/2$, $u = \pi$. Then:

4. Combine the results. …

Method: Definite integral of $\cos^{2}x$ via power reduction

An even power of $\cos$ (or $\sin$) has no elementary antiderivative as-is; use a double-angle identity to reduce the power first, then evaluate.

Steps

Step 1: Apply the power-reduction identity.

Step 2: Integrate the reduced form.

$\int\cos^{2}x\,dx=\frac{x}{2}+\frac{\sin 2x}{4}+C.$ …

Common Mistakes

Mistake 1: Writing $\int\cos^{2}x\,dx=\dfrac{\cos^{3}x}{3}$.

Why it's wrong: that is a chain-rule guess with no $-\sin x$ present; $\cos^{2}x$ is not $u^{2}\,du$. Correct approach: use $\cos^{2}x=\dfrac{1+\cos 2x}{2}$.

Mistake 2: Getting the identity sign wrong ($\cos^{2}x=\dfrac{1-\cos 2x}{2}$).

Why it's wrong: that is $\sin^{2}x$; $\cos^{2}x$ uses a plus. Correct approach: $\cos^{2}x=\dfrac{1+\cos 2x}{2}$. …