Integral x³.sin(tan inverse x⁴)/1+x⁸

We are tasked with evaluating the integral:

\[
I = \int \frac{x^3 \cdot \sin(\tan^{-1}(x^4))}{1 + x^8} \, dx
\]

To solve this, let's break the problem down step-by-step:

### Step 1: Substitution
Let’s first substitute \( u = \tan^{-1}(x^4) \), which simplifies the trigonometric function in the integrand.

This gives:

\[
u = \tan^{-1}(x^4)
\]
\[
\frac{du}{dx} = \frac{4x^3}{1 + x^8}
\]

This substitution simplifies our integral as follows:

\[
du = \frac{4x^3}{1 + x^8} \, dx
\]

Now, rewrite the original integral:

\[
I = \frac{1}{4} \int \sin(u) \, du
\]

### Step 2: Solving the Integral
The integral of \( \sin(u) \) is straightforward:

\[
\int \sin(u) \, du = -\cos(u) + C
\]

So, we have:

\[
I = -\frac{1}{4} \cos(u) + C
\]

### Step 3: Substituting Back
Since we made the substitution \( u = \tan^{-1}(x^4) \), we substitute back to obtain the final solution:

\[
I = -\frac{1}{4} \cos(\tan^{-1}(x^4)) + C
\]

Thus, the solution to the integral is:

\[
I = -\frac{1}{4} \cos(\tan^{-1}(x^4)) + C
\]

---

Would you like further details on any part of the process or clarifications?

Here are 5 related questions that build on this:

1. What are common trigonometric integrals, and how are they solved?
2. How can substitution help simplify complex integrals?
3. What is the derivative of \( \tan^{-1}(x) \), and how is it derived?
4. How can you verify the solution to an integral using differentiation?
5. How do we apply inverse trigonometric functions in calculus?

**Tip:** When faced with complex integrals, substitution is often a powerful tool to simplify the integrand.

Explore the solution to the integral x³.sin(tan⁻¹(x⁴)) / (1+x⁸) using trigonometric substitution and inverse trigonometric functions. This step-by-step guide breaks down complex calculus concepts for advanced high school and undergraduate students.

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