Why does the basic intuition fail here? What is the reason behind it?

Question

We know that $$\int f(ax + b)\,\mathrm dx = \frac1aF(ax + b) + C$$

so while solving $$\int x^2\,\mathrm dx = \frac13x^3 + C$$

also, $$\int (x + 3)^4\,\mathrm dx = \frac15(x + 3)^5 + C$$

but for trig functions like, $$\int \sin^2x\,\mathrm dx \neq \frac13\sin^3x + C$$

Instead, we have to simplify it first algebraically. So why does the intuition fail here?

Update: I seem to be confused with f. Till now $$\int \sin^2(1*x+0)\,\mathrm dx \ $$, I have assumed f = (sin(x))^2 with a=1 and b=0. There seems something wrong with my understanding.

Answer 1

The main reason why you can't apply your formula here is because with your example:

$$\int (\sin x)^2 \,\mathrm dx$$

Here, $f(x) = x^2$ but the input, $\sin x$, does not fit the format of $ax + b$.

More generally, this is because of the chain rule. It states that:

$$\left(f(g(x)\right)' = f'(g(x)) \cdot g'(x)$$

In each of your examples, the chain rule creates a situation where you don't necessarily see it, but it's there. For example when $f(x) = 3x^2$ and $g(x) = x + 3$, we try to find the derivative of $f(g(x))$:

$${\mathrm d \over \mathrm dx} {(x+3)^3\over 3} = 3(x + 3)^2 \cdot {\mathrm d \over \mathrm dx} (x + 3)$$

Since the derivative of $x + 3$ is $1$, you can simply apply the reverse power rule for integration. But when the inside does not have a derivative of $1$ (or a constant for that matter), you run into problems. For example when $f(x) = \frac 13 x^3$ and $g(x) = \sin x$:

$${\mathrm d \over \mathrm dx} \frac 13 (\sin x)^3 = \frac 13 \cdot 3(\sin x)^2 \cdot {\mathrm d \over \mathrm dx} \sin x = \sin^2 x \cos x$$

You'll notice that $\sin^2 x \cos x\neq \sin^2 x$. The chain rule always applies, because you must measure the change in every intermediary function with respect to some variable. See $u$-substitution, which is a technique for integration that essentially undoes the chain rule.

Answer 2

It is not intuition. Your first line is true, $$\int f(ax+b)dx$$ and use the substitution $u=ax+b$ and we get $$\int f(u)du*\frac{1}{a}$$ So far so good. Now if $f(x)=x^3$, then $(x+4)^3=f(x+4)$ hence it fits into our formula and you cna use it. However, if $f(x)=x^2$, then we do not have $a$ and $b$ such that $f(ax+b)=sin(x)^2$ so we cannot use the formula.