Question

Evaluate the indefinite integral: \[ \int \frac{x^3 \sin\left(\tan^{-1}(x^4)\right)}{1+x^8}\, dx \]

• A. \( \frac{1}{4} \cos [\tan^{-1}(x^4)] + C \)
• B. \( -\frac{1}{4} \cos [\tan^{-1}(x^4)] + C \)
• C. \( \frac{1}{4} \sin [\tan^{-1}(x^4)] + C \)
• D. \( -\frac{1}{4} \sin [\tan^{-1}(x^4)] + C \)

Hint:

Whenever you see a composite function like \(f(x^n)\), check if \(x^{n-1}\) appears in the numerator. It is a strong indicator that substitution will simplify the integral immediately.

Answer 1

The Correct Option is B

Concept: This integral is solved using the method of substitution. When a composite function appears along with its derivative structure in the integrand, substitution simplifies the expression. Key formulas:
• Chain rule: \[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \]
• Derivative of inverse tangent: \[ \frac{d}{dx}(\tan^{-1} u) = \frac{1}{1+u^2}\frac{du}{dx} \]
• Integral: \[ \int \sin t \, dt = -\cos t + C \]

Step 1: Identifying substitution.
Observe: \[ \int \frac{x^3 \sin[\tan^{-1}(x^4)]}{1+x^8} dx \] Let: \[ t = \tan^{-1}(x^4) \]

Step 2: Differentiating substitution.
\[ \frac{dt}{dx} = \frac{1}{1+(x^4)^2} \cdot 4x^3 \] \[ \frac{dt}{dx} = \frac{4x^3}{1+x^8} \] So, \[ dt = \frac{4x^3}{1+x^8} dx \] \[ \frac{x^3}{1+x^8} dx = \frac{1}{4} dt \]

Step 3: Substituting into integral.
\[ I = \int \sin(t) \cdot \frac{1}{4} dt \] \[ I = \frac{1}{4} \int \sin t \, dt \]

Step 4: Integration.
\[ I = \frac{1}{4}(-\cos t) + C \] \[ I = -\frac{1}{4}\cos t + C \] Substitute back \(t = \tan^{-1}(x^4)\): \[ I = -\frac{1}{4}\cos[\tan^{-1}(x^4)] + C \]