Q.Integrate the function $\frac{x^3\sin(\tan^{-1}x^4)}{1+x^8}$

The key idea is U Substitution — the numerator contains $x^3$, and the denominator has $1 + x^8$, which suggests a substitution involving $x^4$.

Let $u = x^4$. Then $du = 4x^3 \, dx$, so $x^3 \, dx = \frac{du}{4}$.

Also, $x^8 = (x^4)^2 = u^2$, and $\tan^{-1}(x^4) = \tan^{-1} u$.

The integral becomes:

Now note that $\sin(\tan^{-1} u) = \frac{u}{\sqrt{1+u^2}}$. But here, the denominator $1+u^2$ suggests a simpler path: let $v = \tan^{-1} u$, so $dv = \frac{du}{1+u^2}$. …

Why U-Substitution Works Here

When you see a messy composition like $\sin(\tan^{-1}x^4)$ inside an integral, your first instinct should be: can I make the inside of that sine function my new variable? The presence of $x^3$ in the numerator and $1+x^8$ in the denominator is a dead giveaway — those are exactly the derivative pieces you'd get from differentiating $\tan^{-1}(x^4)$.

Let’s check: if $u = \tan^{-1}(x^4)$, then $du = \frac{4x^3}{1+x^8}\,dx$. That $4x^3/(1+x^8)$ is almost exactly what we have, except we have $x^3/(1+x^8)$ — off by a factor of 4. That’s perfect: a constant factor is easy to fix.

Step-by-Step Solution

1. Set up the substitution. Let $u = \tan^{-1}(x^4)$. Then differentiate:

So $du = \frac{4x^3}{1+x^8}\,dx$.

2. Match the integrand to $du$. Our integrand is $\frac{x^3}{1+x^8} \sin(\tan^{-1}x^4)\,dx$. Notice that $\frac{x^3}{1+x^8}\,dx$ is exactly $\frac{1}{4} du$, because:

And $\sin(\tan^{-1}x^4)$ becomes $\sin u$.

3. Rewrite the integral. …

Method: Substitute the inner argument of a composite (reverse chain rule)

For integrands built as nested functions — for instance $\sin(\tan^{-1}(x^4))$ divided by things — choose $u$ to be the innermost argument whose derivative (up to a constant) is present in the rest of the integrand.

Steps

Step 1: Identify the composite's inner argument.

Look for the "function of a function." Setting $u$ equal to the inner argument is the right move only if $u'$ appears elsewhere in the integrand.

Step 2: Differentiate the candidate and match against the leftover factors.

Test $u=\tan^{-1}(x^4)$: by the chain rule,

The integrand already contains $\dfrac{x^3}{1+x^8}\,dx$, which is $\tfrac14\,du$ — a match up to the constant $\tfrac14$.

Step 3: Rewrite, integrate, and account for the constant. …

Common Mistakes

Mistake 1: Dropping the factor $\tfrac14$ when matching $du$.

Why it's wrong: $du=\dfrac{4x^3}{1+x^8}\,dx$, but the integrand only supplies $\dfrac{x^3}{1+x^8}\,dx=\tfrac14\,du$. Ignoring the $\tfrac14$ scales the whole answer by $4$. Correct approach: divide by the constant, giving $\tfrac14\int\sin u\,du$.

Mistake 2: Substituting $u=x^4$ and getting stuck on $\sin(\tan^{-1}u)$. …