Step 1: Rewrite tan as sin over cos:
Using $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$, rewrite the integral as:
\[ \int x^2 \tan(x^3+2)\,dx = \int x^2 \frac{\sin(x^3+2)}{\cos(x^3+2)}\,dx \]
Step 2: Substitute the denominator:
Let $v = \cos(x^3+2)$. Differentiating, $dv = -3x^2\sin(x^3+2)\,dx$, so $x^2\sin(x^3+2)\,dx = -\dfrac{dv}{3}$.
The integral becomes:
\[ \int \frac{x^2\sin(x^3+2)}{\cos(x^3+2)}\,dx = \int \frac{-dv/3}{v} = -\frac{1}{3}\int \frac{dv}{v} \]
Step 3: Integrate and simplify:
Using $\int \dfrac{dv}{v} = \ln|v| + C$:
\[ -\frac{1}{3}\ln|v| + C = -\frac{1}{3}\ln|\cos(x^3+2)| + C \]
Since $-\ln|\cos\theta| = \ln|\sec\theta|$, this simplifies to the same form as before.
Final Answer:
Substituting for cosine instead of the whole angle argument gives the identical result.
\[ \boxed{\int x^2\tan(x^3+2)\,dx = \frac{1}{3}\ln\left|\sec(x^3+2)\right| + C} \]