Question

$\int \frac{x^4 \cos(\tan^{-1} x^5)}{1 + x^{10}} dx$ equals ______.

• A. $\sin(\tan^{-1} x^5) + c$
• B. $x^4 \sin(\tan^{-1} x^5) + c$
• C. $\frac{1}{5}\sin(\tan^{-1} x^5) + c$
• D. $\cos(\tan^{-1} x^5) + c$

Hint:

Always look for the function-derivative pair in integration problems. If you see $f(g(x))$ multiplied by something that strongly resembles $g'(x)$, substitution $t = g(x)$ will almost certainly solve it.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are tasked with solving an indefinite integral involving a trigonometric function composed with an inverse trigonometric function and an algebraic fraction. This heavily suggests the method of substitution.

Step 2: Key Formula or Approach:
Identify a suitable substitution. Notice that the derivative of $\tan^{-1}(u)$ is $\frac{1}{1+u^2} \cdot du$.
If we let $t = \tan^{-1}(x^5)$, we can find $dt$ and see if it perfectly matches the rest of the integrand.

Step 3: Detailed Explanation:
Let $t = \tan^{-1}(x^5)$.
Differentiate both sides with respect to $x$ using the chain rule:
$$\frac{dt}{dx} = \frac{1}{1 + (x^5)^2} \cdot \frac{d}{dx}(x^5)$$
$$\frac{dt}{dx} = \frac{1}{1 + x^{10}} \cdot 5x^4$$
Rearranging to solve for the differential $dx$:
$$\frac{dt}{5} = \frac{x^4}{1 + x^{10}} dx$$
Notice that the term $\frac{x^4}{1 + x^{10}} dx$ is exactly present in the original integral!
Substitute $t$ and $dt/5$ back into the integral:
$$\int \cos(t) \cdot \frac{dt}{5} = \frac{1}{5} \int \cos(t) \, dt$$
Integrate with respect to $t$:
$$= \frac{1}{5} \sin(t) + c$$
Finally, back-substitute the original expression for $t$:
$$= \frac{1}{5} \sin(\tan^{-1}(x^5)) + c$$

Step 4: Final Answer:
The integrated expression matches option (c).