Find the integral of x^3 sin(7x^4 + 3) dx.

Step-by-step solution

1. 1

   Choose substitution over integration by parts. By parts is the reflex for a product, but here the outside factor $x^{3}$ is (up to a constant) exactly the derivative of the inside of the sine. That is the signature of a $u$-substitution, and by parts would only make the integral worse.

2. 2

   Set up the substitution. Let

   $u=7x^{4}+3\qquad\Longrightarrow\qquad du=28x^{3}\,dx\qquad\Longrightarrow\qquad x^{3}\,dx=\frac{du}{28}$

   The constant $\tfrac1{28}$ is what absorbs the mismatch between $x^{3}$ and $28x^{3}$.

3. 3

   Rewrite the integral entirely in $u$.

   $\int x^{3}\sin\!\left(7x^{4}+3\right)dx=\frac{1}{28}\int\sin u\,du$

   No $x$ remains, which is the check that the substitution was the right one.

4. 4

   Integrate and substitute back. Since $\displaystyle\int\sin u\,du=-\cos u$:

   $\frac{1}{28}\left(-\cos u\right)+C=-\frac{\cos\!\left(7x^{4}+3\right)}{28}+C$

   The minus sign comes from the antiderivative of sine and is the easiest thing to drop here.

5. 5

   Verify by differentiating back. By the chain rule,

   $\frac{d}{dx}\left[-\frac{\cos\!\left(7x^{4}+3\right)}{28}\right]=\frac{\sin\!\left(7x^{4}+3\right)\cdot28x^{3}}{28}=x^{3}\sin\!\left(7x^{4}+3\right)\ \checkmark$

   A numerical derivative confirms this at $x=0.4$, $0.9$ and $1.3$ to within $10^{-5}$ ✓. The $28$ cancels exactly, which is the whole reason the constant was chosen that way.