Q.Find an anti derivative (or integral) of the function $\sin 2x - 4 e^{3x}$ by the method of inspection.

Concept: Power Rule Integration (reverse differentiation) — we look for a function whose derivative gives the given expression.

Step 1: For $\sin 2x$, recall $\frac{d}{dx}(\cos 2x) = -2\sin 2x$. So $\frac{d}{dx}\left(-\frac{1}{2}\cos 2x\right) = \sin 2x$.

Step 2: For $-4e^{3x}$, recall $\frac{d}{dx}(e^{3x}) = 3e^{3x}$. So $\frac{d}{dx}\left(-\frac{4}{3}e^{3x}\right) = -4e^{3x}$. …

The method of inspection means you look at a function and ask: "What function, when differentiated, gives me this?" It's the opposite of differentiation — you're essentially doing reverse differentiation using your memory of standard derivatives.

Here, we have two separate terms: $\sin 2x$ and $-4e^{3x}$. Let's handle them one at a time.

1. For the $\sin 2x$ term: Recall that $\frac{d}{dx}(\cos x) = -\sin x$. More generally, $\frac{d}{dx}(\cos 2x) = -2\sin 2x$ (by the chain rule). We want $\sin 2x$, not $-2\sin 2x$. So if $\frac{d}{dx}(\cos 2x) = -2\sin 2x$, then dividing both sides by $-2$ gives:

So one antiderivative of $\sin 2x$ is $-\frac{1}{2}\cos 2x$.

2. For the $-4e^{3x}$ term:

   We know $\frac{d}{dx}(e^{3x}) = 3e^{3x}$. We have $-4e^{3x}$, which is $-\frac{4}{3}$ times the derivative of $e^{3x}$.

   Check: $\frac{d}{dx}\left(-\frac{4}{3}e^{3x}\right) = -\frac{4}{3} \cdot 3e^{3x} = -4e^{3x}$.

   So one antiderivative of $-4e^{3x}$ is $-\frac{4}{3}e^{3x}$.

3. Combine the two results:

   Since differentiation is linear (the derivative of a sum is the sum of derivatives), the antiderivative of the sum is the sum of the antiderivatives.

   Therefore, an antiderivative of $\sin 2x - 4e^{3x}$ is:

4. Don't forget the constant: …

Method: Antiderivative by inspection

"By inspection" means recognising each term as the derivative of a known function and reversing it, adjusting the constant factor.

Steps

Step 1: Recall the matching derivative.

$\dfrac{d}{dx}\cos kx=-k\sin kx$ and $\dfrac{d}{dx}e^{kx}=ke^{kx}$.

Step 2: Adjust the constant. …

Common Mistakes

Mistake 1: Taking $-\cos 2x$ as the antiderivative of $\sin 2x$.

Why it's wrong: $\frac{d}{dx}(-\cos 2x)=2\sin 2x$, a factor of $2$ too big. Correct approach: the antiderivative is $-\frac12\cos 2x$.

Mistake 2: Integrating $e^{3x}$ as $e^{3x}$ or $\frac{1}{3x}e^{3x}$. …