Q.Evaluate the definite integral: $\int_0^{\pi/4} \sin 2x \ dx$

The key idea is that this is a straightforward definite integral of a trigonometric function — no symmetry trick needed here, just direct integration.

Step 1: Recall the antiderivative.

$\int \sin 2x \, dx = -\frac{1}{2} \cos 2x + C$.

Step 2: Apply the limits $0$ to $\frac{\pi}{4}$:

$\left[-\frac{1}{2} \cos 2x \right]_0^{\pi/4} = -\frac{1}{2} \cos\left(\frac{\pi}{2}\right) - \left(-\frac{1}{2} \cos 0\right)$. …

The key idea here is that the integrand $\sin 2x$ is a scaled version of the basic sine function. When you see an argument like $2x$, your first instinct should be to think about how the chain rule works in reverse — that is, substitution.

Let’s walk through it.

1. Recognize the form.

   The integral is $\int \sin(2x) \, dx$. If it were just $\int \sin x \, dx$, the answer would be $-\cos x + C$. But because the argument is $2x$, the derivative of $2x$ (which is $2$) will appear when we differentiate $\cos(2x)$. So the antiderivative will involve a factor of $\frac{1}{2}$.

2. Use substitution (or pattern recall).

   Let $u = 2x$. Then $du = 2 \, dx$, so $dx = \frac{du}{2}$.

   When $x = 0$, $u = 0$. When $x = \frac{\pi}{4}$, $u = \frac{\pi}{2}$.

   The integral becomes:

3. Evaluate the simpler integral. The antiderivative of $\sin u$ is $-\cos u$. So:

We know $\cos\frac{\pi}{2} = 0$ and $\cos 0 = 1$. So this becomes:

$\frac{1}{2} \left( -0 + 1 \right) = \frac{1}{2}.$ …

Method: Definite integral of $\sin 2x$ (constant-multiple substitution)

Integrate $\sin(kx)$ to $-\dfrac{\cos(kx)}{k}$, then evaluate across the limits.

Steps

Step 1: Recall $\int\sin(kx)\,dx=-\dfrac{\cos(kx)}{k}$. Here $k=2$, so $F(x)=-\dfrac{\cos 2x}{2}$.

Step 2: Set the bracket. …

Common Mistakes

Mistake 1: Integrating $\sin 2x$ to $-\cos 2x$ (missing the $\tfrac12$).

Why it's wrong: the inner factor $2$ demands dividing by $2$: $\int\sin 2x\,dx=-\dfrac{\cos 2x}{2}$. Correct approach: divide by $k$.

Mistake 2: Sign error on $\int\sin$. …