Question

\(\int\frac{x^3\sin(\tan^{-1}(x^4))}{1+x^8}dx\) is equal to

• A. \(\frac{-\cos(\tan^{-1}(x^4))}{4}+C\)
• B. \(\frac{\cos(\tan^{-1}(x^4))}{4}+C\)
• C. \(\frac{-\cos(\tan^{-1}(x^3))}{3}+C\)
• D. \(\frac{-\sin(\tan^{-1}(x^4))}{4}+C\)

Answer 1

The Correct Option is A

Let's solve the integral \( \int \frac{x^3 \sin (\tan^{-1}(x^4))}{1 + x^8} \, dx \). We use substitution to solve this integral. Let: \[ u = \tan^{-1}(x^4) \] Differentiating both sides with respect to \( x \), we get: \[ \frac{du}{dx} = \frac{4x^3}{1 + x^8} \] This implies: \[ dx = \frac{1 + x^8}{4x^3} du \] Substituting into the integral, we get: \[ \int \frac{x^3 \sin(\tan^{-1}(x^4))}{1 + x^8} \, dx = \frac{1}{4} \int \sin(u) \, du \] The integral of \( \sin(u) \) is \( -\cos(u) \), so: \[ = - \frac{\cos(u)}{4} + C \] Substituting back \( u = \tan^{-1}(x^4) \), we get: \[ = - \frac{\cos(\tan^{-1}(x^4))}{4} + C \] Thus, the correct answer is: \[{- \frac{\cos(\tan^{-1}(x^4))}{4} + C} \]

The correct answer is (A) : \(\frac{-\cos(\tan^{-1}(x^4))}{4}+C\).

Answer 2

Let's evaluate the integral: \( \int\frac{x^3\sin(\tan^{-1}(x^4))}{1+x^8}dx \)

Let \( u = \tan^{-1}(x^4) \). Then, \( \frac{du}{dx} = \frac{1}{1 + (x^4)^2} \cdot 4x^3 = \frac{4x^3}{1 + x^8} \).

Therefore, \( du = \frac{4x^3}{1 + x^8} dx \), which implies \( \frac{1}{4} du = \frac{x^3}{1 + x^8} dx \).

Substituting this into the integral, we get:

\( \int \sin(u) \cdot \frac{1}{4} du = \frac{1}{4} \int \sin(u) du \)

The integral of \( \sin(u) \) is \( -\cos(u) \), so we have:

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

Now, substitute back \( u = \tan^{-1}(x^4) \):

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

Thus, the integral is equal to \( \frac{-\cos(\tan^{-1}(x^4))}{4}+C \).