Q.Write an anti derivative for each of the following functions using the method of inspection:

Concept: Power Rule Integration — we reverse the derivative by asking: which function, when differentiated, gives the given expression?

(i) $\cos 2x$

We know $\frac{d}{dx} \sin 2x = 2\cos 2x$. To get $\cos 2x$, we multiply by $\frac12$:

$\frac{d}{dx}\left( \frac12 \sin 2x \right) = \cos 2x$.

So an antiderivative is $\frac12 \sin 2x$.

(ii) $3x^2 + 4x^3$

Differentiate $x^3$ to get $3x^2$, and $x^4$ to get $4x^3$.

Thus $\frac{d}{dx}(x^3 + x^4) = 3x^2 + 4x^3$.

So an antiderivative is $x^3 + x^4$.

(iii) $\dfrac{1}{x},\ x \neq 0$

We know $\frac{d}{dx} \log|x| = \frac{1}{x}$.

So an antiderivative is $\log|x|$.

The idea behind "method of inspection" is simple: you look at the given function and ask yourself, "What function, when differentiated, gives me this?" It's reverse differentiation — you rely on your memory of standard derivatives and then adjust for constants. This is the most intuitive way to find antiderivatives, especially for simple functions.

Let's work through each one.

1. (i) $\cos 2x$

We know that $\frac{d}{dx}(\sin x) = \cos x$. But here the argument is $2x$, not $x$. So we need a function whose derivative brings out a factor of 2 from the chain rule.

Think: $\frac{d}{dx}(\sin 2x) = \cos 2x \cdot 2 = 2\cos 2x$. That gives us $2\cos 2x$, but we want just $\cos 2x$. So we need to divide by 2 to cancel the extra factor.

Therefore, $\frac{d}{dx}\left(\frac{1}{2}\sin 2x\right) = \frac{1}{2} \cdot 2\cos 2x = \cos 2x$.

So the antiderivative is $\frac{1}{2}\sin 2x + C$, where $C$ is any constant (since derivative of a constant is zero).

2. (ii) $3x^2 + 4x^3$

This is a sum of two power functions. The power rule for differentiation says $\frac{d}{dx}(x^n) = nx^{n-1}$. For antiderivatives, we reverse this: if the derivative gives $nx^{n-1}$, then the antiderivative of $x^{n-1}$ is $\frac{x^n}{n}$ (for $n \neq 0$).

Let's handle each term separately.

For $3x^2$: We need a function whose derivative is $x^2$. Since $\frac{d}{dx}(x^3) = 3x^2$, we have exactly $3x^2$ as the derivative of $x^3$. So the antiderivative of $3x^2$ is $x^3$.

For $4x^3$: We need a function whose derivative is $x^3$. Since $\frac{d}{dx}(x^4) = 4x^3$, the antiderivative of $4x^3$ is $x^4$.

Adding them together: the antiderivative of $3x^2 + 4x^3$ is $x^3 + x^4 + C$.

3. (iii) $\frac{1}{x},\ x \neq 0$

This is the special case where the power rule fails (since $n = -1$ would give division by zero). We need a function whose derivative is $\frac{1}{x}$.

From standard derivatives, we know $\frac{d}{dx}(\log x) = \frac{1}{x}$ for $x > 0$. But the domain here is $x \neq 0$, which includes negative $x$ as well. For $x < 0$, $\log x$ is not defined, but $\log(-x)$ works. The clean way to handle both positive and negative $x$ is to use $\log|x|$.

Check: $\frac{d}{dx}(\log|x|) = \frac{1}{x}$ for all $x \neq 0$. (For $x > 0$, it's $\log x$; for $x < 0$, it's $\log(-x)$, whose derivative is $\frac{1}{-x} \cdot (-1) = \frac{1}{x}$.)

So the antiderivative is $\log|x| + C$.

Method: Antiderivative by Inspection (Reverse Differentiation)

Use this when the integrand is a standard function (a power, an exponential, a basic trig function, or a simple constant multiple of one) whose antiderivative you can recognise by asking "what did I differentiate to get this?"

Steps

Step 1: Recall the matching standard derivative.

Scan your table of standard results and find the function whose derivative has the same shape as the integrand. For example, $\dfrac{d}{dx}(\sin kx)=k\cos kx$, $\dfrac{d}{dx}(x^{n+1})=(n+1)x^n$, and $\dfrac{d}{dx}(\log|x|)=\dfrac1x$.

Step 2: Fix the constant multiple.

Differentiating a composite like $\sin kx$ pulls out a factor $k$ (chain rule). To cancel it, divide your guess by that constant:

For a pure power, divide by the new exponent: $\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C$ (valid for $n\neq -1$).

Step 3: Handle the exceptional case $n=-1$.

The power rule fails for $\dfrac1x$; instead use $\int \dfrac1x\,dx=\log|x|+C$.

Step 4: Verify and add $C$.

Differentiate your answer mentally to confirm it returns the integrand, and always attach the arbitrary constant $C$ because antiderivatives are unique only up to a constant.

Common Mistakes

Mistake 1: Writing $\int \cos 2x\,dx = \sin 2x + C$ without the $\tfrac12$.

Why it's wrong: differentiating $\sin 2x$ gives $2\cos 2x$, not $\cos 2x$, because of the chain-rule factor of $2$. Correct approach: divide by the inner derivative, $\int\cos 2x\,dx=\tfrac12\sin 2x+C$.

Mistake 2: Applying the power rule to $\dfrac1x$.

Why it's wrong: the rule $\int x^n\,dx=\frac{x^{n+1}}{n+1}$ blows up at $n=-1$ (division by zero). Correct approach: $\int\frac1x\,dx=\log|x|+C$, with the modulus so it is valid for $x<0$ too.

Mistake 3: Dropping the constant of integration.

Why it's wrong: every function of the form $F(x)+C$ is an antiderivative, so an answer without $C$ is incomplete. Correct approach: always append $+C$.