`int(x^(3)sin[tan^(-1)(x^(4))])/(1+x^(8))dx=`

Options

A. `1/4cos[tan^(-1)(x^(4))]+c`
B. `1/4sin[tan^(-1)(x^(4))]+c`
C. `(-1)/4cos[tan^(-1)(x^(4))]+c`
D. `1/4sec[tan^(-1)(x^(4))]+c`

Answer

To solve the integral

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

we will use substitution and integration techniques.

### Step 1: Substitution
Let \( u = \tan^{-1}(x^4) \). Then, we differentiate \( u \) with respect to \( x \):

\[ \frac{du}{dx} = \frac{1}{1 + (x^4)^2} \cdot 4x^3 = \frac{4x^3}{1 + x^8}. \]

This implies that

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

### Step 2: Rewrite the Integral
Now, we can rewrite the integral in terms of \( u \):

\[ \int \frac{x^3 \sin(u)}{1+x^8} \cdot \frac{1 + x^8}{4x^3} \, du = \int \frac{\sin(u)}{4} \, du. \]

### Step 3: Integrate
Now we can integrate:

\[ \int \frac{\sin(u)}{4} \, du = \frac{1}{4} \int \sin(u) \, du = -\frac{1}{4} \cos(u) + C. \]

### Step 4: Substitute Back
Now substitute back \( u = \tan^{-1}(x^4) \):

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

### Step 5: Simplify \( \cos(\tan^{-1}(x^4)) \)
Recall that if \( u = \tan^{-1}(x^4) \), then \( \tan(u) = x^4 \). We can use the identity:

\[ \cos(u) = \frac{1}{\sqrt{1 + \tan^2(u)}} = \frac{1}{\sqrt{1 + (x^4)^2}} = \frac{1}{\sqrt{1 + x^8}}. \]

Thus, our final answer is:

\[ -\frac{1}{4} \cdot \frac{1}{\sqrt{1 + x^8}} + C. \]

### Final Answer
\[ \int \frac{x^3 \sin(\tan^{-1}(x^4))}{1+x^8} \, dx = -\frac{1}{4\sqrt{1 + x^8}} + C. \]