Q.Integrate the following function: $\sin 2x$

The key idea is to use the sine double-angle identity to rewrite $\sin 2x$ in a form that integrates directly.

Step 1: Recall the identity $\sin 2x = 2 \sin x \cos x$.

Step 2: Integrate term by term:

Step 3: Use substitution $u = \sin x$, $du = \cos x \, dx$, giving

Alternatively, integrate directly: $\int \sin 2x \, dx = -\frac{1}{2} \cos 2x + C$, which is equivalent.

The key insight here is that $\sin 2x$ is not a basic integral we memorize directly — but it is a simple transformation of a basic one. The sine double-angle formula tells us $\sin 2x = 2\sin x \cos x$, which might look more complicated. Instead, the cleanest approach is to notice that the derivative of $\cos 2x$ is $-2\sin 2x$, so the antiderivative of $\sin 2x$ must be $-\frac{1}{2}\cos 2x$.

Let’s work through it step by step.

1. Recognize the pattern.

   We know that $\frac{d}{dx}(\cos 2x) = -2\sin 2x$ by the chain rule. This tells us that $\sin 2x$ is almost the derivative of $\cos 2x$, except for a factor of $-2$.

2. Set up the integral.

   We want $\int \sin 2x \, dx$. If $\frac{d}{dx}(\cos 2x) = -2\sin 2x$, then dividing both sides by $-2$ gives:

3. Write the antiderivative. Therefore,

where $C$ is the constant of integration.

If you prefer substitution, let $u = 2x$, then $du = 2\,dx$, so $dx = \frac{du}{2}$. The integral becomes $\int \sin u \cdot \frac{du}{2} = \frac{1}{2}\int \sin u \, du = -\frac{1}{2}\cos u + C = -\frac{1}{2}\cos 2x + C$. Same result.

Method: Integrating $\sin(kx)$ and other $\sin/\cos$ of a linear argument

Use this for any $\int \sin(kx)\,dx$ or $\int \cos(kx)\,dx$ where the angle is a constant times $x$. The only new ingredient beyond the basic sine/cosine integrals is a compensation factor $\tfrac1k$.

Steps

Step 1: Recall the basic antiderivative and why $k$ appears.

Because $\dfrac{d}{dx}\cos(kx)=-k\sin(kx)$, undoing it needs a $-\tfrac1k$:

Step 2: Identify $k$ from the argument.

Read off the multiplier of $x$ inside the trig function (here $k=2$). This single number is the compensation factor.

Step 3: Write the antiderivative with the $\tfrac1k$ factor.

Step 4: Verify by differentiating.

Differentiate your answer; the chain rule should regenerate exactly the integrand. (Equivalently, substitute $u=kx$, $du=k\,dx$, to see the $\tfrac1k$ emerge.) This check catches the near-universal error of omitting $\tfrac1k$.

Common Mistakes

Mistake 1: Omitting the $\tfrac1k$ factor.

Why it's wrong: writing $\int\sin 2x\,dx=-\cos 2x+C$ differentiates back to $2\sin 2x$, not $\sin 2x$. Correct approach: include the compensation factor, giving $-\tfrac12\cos 2x+C$.

Mistake 2: Sign error on the cosine.

Why it's wrong: $\int\sin(kx)\,dx$ is negative cosine; students sometimes write $+\tfrac12\cos 2x$. Correct approach: remember $\int\sin=-\cos$, then differentiate to confirm the sign.

Mistake 3: Treating $\sin 2x=2\sin x\cos x$ as harder.

Why it's wrong: expanding is fine but tempts errors; both $-\tfrac12\cos 2x+C$ and $\sin^2 x+C$ are correct and differ only by a constant. Correct approach: use the direct $\tfrac1k$ rule, or if expanding, accept the equivalent $\sin^2 x$ form.