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

The key idea is to use the symmetry of $\cos^2 x$ over $[0, \pi/2]$, or equivalently, the identity $\cos^2 x = 1 - \sin^2 x$ combined with the property $\int_0^{\pi/2} f(\sin x)\,dx = \int_0^{\pi/2} f(\cos x)\,dx$.

Let $I = \int_0^{\pi/2} \cos^2 x \, dx$. Using the property $\int_0^{\pi/2} f(\sin x)\,dx = \int_0^{\pi/2} f(\cos x)\,dx$, we also have $I = \int_0^{\pi/2} \sin^2 x \, dx$.

Adding the two expressions:

Thus $I = \frac{\pi}{4}$.

The problem asks us to evaluate $\int_{0}^{\pi/2} \cos^2 x \, dx$ using properties of definite integrals. The direct approach — finding an antiderivative — is straightforward, but the instruction to use properties nudges us toward a more elegant method that builds deeper intuition.

The key property here is the symmetry of the definite integral about the midpoint of the interval. For any function $f$ continuous on $[0, a]$, we have:

Why does this work? Because as $x$ runs from $0$ to $a$, the quantity $a - x$ runs from $a$ down to $0$ — it’s just a reversal of direction. The area under the curve doesn’t care about direction, so the integral stays the same.

Now, apply this to our integral. Let:

Here $a = \frac{\pi}{2}$. Using the property:

But $\cos\left(\frac{\pi}{2} - x\right) = \sin x$, so:

This is the crucial step: the integral of $\cos^2 x$ from $0$ to $\pi/2$ equals the integral of $\sin^2 x$ over the same interval.

Now add the two expressions for $I$:

And $\cos^2 x + \sin^2 x = 1$, the most fundamental identity in trigonometry. So:

The integral of $1$ from $0$ to $\pi/2$ is just the length of the interval: $\frac{\pi}{2} - 0 = \frac{\pi}{2}$.

Thus:

Method: The reflection property $\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$

Replacing $x$ by $a-x$ leaves a definite integral over $[0,a]$ unchanged. Adding the original and reflected forms often produces a trivially integrable sum.

Steps

Step 1: Name the integral and reflect.

Let $I=\displaystyle\int_0^a f(x)\,dx$. Apply

Step 2: Simplify the reflected integrand.

Use the relevant co-function identities (over $[0,\tfrac{\pi}{2}]$, $\sin(\tfrac{\pi}{2}-x)=\cos x$ and vice-versa), which typically swaps the roles of the functions.

Step 3: Add the two expressions for $I$.

$2I=\displaystyle\int_0^a\big[f(x)+f(a-x)\big]\,dx$; choose the reflection so this sum collapses (e.g. to $1$).

Step 4: Integrate the simple sum and halve.

Solve $2I=\int_0^a(\text{simple})\,dx$ for $I$.

Common Mistakes

Mistake 1: Applying $\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$ with mismatched limits.

Why it's wrong: the property needs a lower limit of $0$ and the same upper limit $a$ inside $f(a-x)$; using it on, say, $\int_0^\pi\cos^2x\,dx$ (where the midpoint differs) gives a wrong reflection. Correct approach: confirm the limits are $0$ to $a$ before reflecting.

Mistake 2: Forgetting that $\cos(\tfrac{\pi}{2}-x)=\sin x$, so $\cos^2$ becomes $\sin^2$.

Why it's wrong: the whole trick relies on the reflected integrand becoming $\sin^2x$ so that $\cos^2x+\sin^2x=1$. Correct approach: use the co-function identity, add, and get $2I=\tfrac{\pi}{2}$.