Q.Evaluate $\int_{-\pi/4}^{\pi/4} \sin^2 x\, dx$

The key idea is that $\sin^2 x$ is an even function, so we can simplify the integral by symmetry.

Step 1: Since $\sin^2(-x) = \sin^2 x$, the integrand is even. Therefore,

Step 2: Use the identity $\sin^2 x = \frac{1 - \cos 2x}{2}$:

Step 3: Integrate:

This problem is a classic example of evaluating a definite integral of a squared trigonometric function. The direct approach — integrating $\sin^2 x$ as is — is messy because the antiderivative isn't immediately obvious. Instead, we use a power-reduction identity to rewrite the integrand into a sum of simpler terms.

The key insight: $\sin^2 x$ oscillates between 0 and 1, and its average value over a full period is $1/2$. The identity $\sin^2 x = \frac{1-\cos 2x}{2}$ converts the squared sine into a constant plus a cosine wave with double frequency. Integrating a constant is trivial, and integrating $\cos 2x$ is straightforward. The limits are symmetric about zero, which also simplifies things slightly, but the method works for any limits.

Let’s work through it step by step.

1. Apply the power-reduction identity. We rewrite the integrand using the standard identity:

This is derived from the double-angle formula $\cos 2x = 1 - 2\sin^2 x$.

The integral becomes:

2. Split the integral into two simpler ones. The constant factor $1/2$ can be pulled out, and the difference inside the integral splits:

3. Evaluate the first integral. The integral of $1$ over an interval is just the length of the interval:

So the first term is:

4. Evaluate the second integral. The antiderivative of $\cos 2x$ is $\frac{\sin 2x}{2}$. Thus:

Compute at the upper limit: $\sin\left(2 \cdot \frac{\pi}{4}\right) = \sin\left(\frac{\pi}{2}\right) = 1$.

At the lower limit: $\sin\left(2 \cdot -\frac{\pi}{4}\right) = \sin\left(-\frac{\pi}{2}\right) = -1$.

So:

Therefore:

5. Combine the results. The second term in step 2 is $-\frac{1}{2} \cdot 1 = -\frac{1}{2}$. Adding the first term:

Method: Power-Reduction Identity for a Definite Integral of $\sin^2$/$\cos^2$

Use this when integrating $\sin^2 x$ or $\cos^2 x$ over given limits: replace the square with a double-angle form that integrates directly.

Steps

Step 1: Apply the power-reduction identity.

This turns the un-integrable square into a constant plus a $\cos 2x$ term.

Step 2: Integrate term-by-term.

Step 3: Apply the limits.

Substitute the bounds and subtract. Using symmetry ($\sin^2x$ is even) can halve the work over $[-c,c]$. For $\left[-\frac\pi4,\frac\pi4\right]$ the value is $\frac\pi4-\frac12$.

Common Mistakes

Mistake 1: Trying to integrate $\sin^2 x$ directly as $\dfrac{\sin^3 x}{3}$.

Why it's wrong: that is not the antiderivative of $\sin^2 x$. Correct approach: use $\sin^2x=\frac{1-\cos2x}{2}$ first.

Mistake 2: Forgetting the $\dfrac14$ factor on $\int\cos 2x\,dx$.

Why it's wrong: $\int\cos2x\,dx=\frac12\sin2x$, and the outer $\frac12$ makes it $\frac14\sin2x$. Correct approach: carry both constants.

Mistake 3: Mis-evaluating $\sin 2x$ at the limits.

Why it's wrong: at $x=\pm\frac\pi4$, $\sin2x=\pm1$; a sign slip changes the result. Correct approach: evaluate $\sin\left(\frac\pi2\right)=1$ and $\sin\left(-\frac\pi2\right)=-1$ carefully.