Step 1 : Understanding the Question:
This problem requires evaluating an indefinite integral that appears complex due to the nested functions: a polynomial $x^3$, a trigonometric sine function, and an inverse tangent function with an $x^4$ argument. To solve such an integral, we look for a functional dependency. Notice that $x^8$ is $(x^4)^2$ and the derivative of $x^4$ is related to $x^3$. This structure is a strong indicator that the "Method of Substitution" will reduce the complexity significantly.
Step 2 : Key Formulas and approach:
We will use the substitution $u = \tan^{-1}(x^4)$. This is strategic because the derivative of $\tan^{-1}(v)$ is $1/(1+v^2)$. By applying the chain rule, we can find a expression for $du$ that perfectly matches the other components of the integrand.
Key formulas:
1. $\frac{d}{dx}[\tan^{-1}(f(x))] = \frac{1}{1 + [f(x)]^2} \cdot f'(x)$
2. $\int \sin u \, du = -\cos u + C$
Our approach is to transform the $x$-based integral into a simple trigonometric $u$-based integral.
Step 3 : Detailed Explanation:
Let $u = \tan^{-1}(x^4)$.
Differentiating $u$ with respect to $x$ using the chain rule: $\frac{du}{dx} = \frac{1}{1 + (x^4)^2} \cdot (4x^3) = \frac{4x^3}{1 + x^8}$.
Rearranging the differential terms: $du = \frac{4x^3}{1 + x^8} dx$, which implies that $\frac{1}{4} du = \frac{x^3}{1 + x^8} dx$.
Now, substitute these parts back into the original integral: $I = \int \sin(u) \cdot \frac{1}{4} du$.
Pull out the constant: $I = \frac{1}{4} \int \sin u \, du$.
Perform the standard integration: $I = \frac{1}{4} (-\cos u) + C = -\frac{1}{4} \cos u + C$.
Finally, replace $u$ with the original substitution $\tan^{-1}(x^4)$ to obtain the result in terms of $x$: $I = -\frac{1}{4} \cos [\tan^{-1}(x^4)] + C$.
Step 4 : Final Answer:
The indefinite integral is evaluated as $-\frac{1}{4} \cos [\tan^{-1}(x^4)] + C$.