We can make two algebraic observations regarding the integrand, \(x^3 \cdot \sin (7x^4 + 3)\text{.}\) First, \(\sin (7x^4 + 3)\) is a composite function; as such, we know we’ll need a more sophisticated approach to antidifferentiating. Second, \(x^3\) is almost the derivative of \((7x^4 + 3)\text{;}\) the only issue is a missing constant. Thus, \(x^3\) and \((7x^4 + 3)\) are nearly a function-derivative pair. Furthermore, we know the antiderivative of \(f(u) = \sin(u)\text{.}\) The combination of these observations suggests that we can evaluate the given indefinite integral by reversing the chain rule through \(u\)-substitution.

Letting \(u\) represent the inner function of the composite function \(\sin (7x^4 + 3)\text{,}\) we have \(u = 7x^4 + 3\text{,}\) and thus \(\frac{du}{dx} = 28x^3\text{.}\) In differential notation, it follows that \(du = 28x^3 \, dx\text{,}\) and thus \(x^3 \, dx = \frac{1}{28} \, du\text{.}\) The original indefinite integral may be slightly rewritten as

\begin{equation*}
\int \sin (7x^4 + 3) \cdot x^3 \, dx\text{,}
\end{equation*}

and so by substituting \(u\) for \(7x^4 + 3\) and \(\frac{1}{28} \, du\) for \(x^3 \, dx\text{,}\) it follows that

\begin{equation*}
\int \sin (7x^4 + 3) \cdot x^3 \, dx = \int \sin(u) \cdot \frac{1}{28} \, du\text{.}
\end{equation*}

Now we may evaluate the easier integral in \(u\text{,}\) and then replace \(u\) by the expression \(7x^4 + 3\text{.}\) Doing so, we find

\begin{align*}
\int \sin (7x^4 + 3) \cdot x^3 \, dx &= \int \sin(u) \cdot \frac{1}{28} \, du\\
&= \frac{1}{28} \int \sin(u) \, du\\
&= \frac{1}{28} (-\cos(u)) + C\\
&= -\frac{1}{28} \cos(7x^4 + 3) + C\text{.}
\end{align*}

To check our work, we observe by the Chain Rule that

\begin{equation*}
\frac{d}{dx} \left[ -\frac{1}{28}\cos(7x^4 + 3) \right] = -\frac{1}{28} \cdot (-1)\sin(7x^4 + 3) \cdot 28x^3 = \sin(7x^4 + 3) \cdot x^3\text{,}
\end{equation*}

which is indeed the original integrand.