Q.Integrate the function $\frac{x^3}{\sqrt{1-x^8}}$

Concept: U Substitution — the denominator’s structure suggests a perfect square inside a square root, so we set $u = x^4$.

Step 1: Let $u = x^4$. Then $du = 4x^3\,dx$, so $x^3\,dx = \frac{du}{4}$.

Step 2: Also $x^8 = (x^4)^2 = u^2$. The integral becomes

Step 3: This is a standard form: $\int \frac{du}{\sqrt{1-u^2}} = \sin^{-1} u + C$. Therefore …

The key insight here is that the denominator contains $\sqrt{1 - x^8}$, and $x^8 = (x^4)^2$. That square inside a square root under $1 - (\text{something})^2$ is a dead giveaway for the arcsine derivative formula: $\frac{d}{du} \sin^{-1} u = \frac{1}{\sqrt{1-u^2}}$.

But we have $x^3$ in the numerator, not $x^4$ or something that directly matches. That’s where substitution comes in — we need to turn the numerator into the derivative of the “something” we want to put inside the arcsine.

1. Choose the substitution.

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

   Why $x^4$? Because $x^8 = (x^4)^2 = u^2$, and the numerator $x^3$ is exactly the derivative of $x^4$ up to a constant factor. This is the cleanest way to match the arcsine form.

2. Rewrite the integral.

   The original integral is

Substituting $u = x^4$ and $x^3\,dx = \frac{du}{4}$ gives

3. Recognise the standard integral.

   The integral $\int \frac{du}{\sqrt{1-u^2}}$ is exactly $\sin^{-1} u + C$. This is a fundamental result from differentiation: $\frac{d}{du} \sin^{-1} u = \frac{1}{\sqrt{1-u^2}}$.

4. Back-substitute.

   So we have

   $\frac{1}{4} \sin^{-1} u + C = \frac{1}{4} \sin^{-1}(x^4) + C.$ …

Method: Substitution Producing an Inverse-Sine Standard Form

Use this when the integrand looks like $\dfrac{(\text{power})}{\sqrt{1-(\text{power})^2}}$: a substitution turns the numerator into $du$ and reveals the arcsine derivative.

Steps

Step 1: Identify the "something-squared" under the root.

Here $1-x^8=1-(x^4)^2$, so the natural inner function is $u=x^4$.

Step 2: Match the numerator to $du$.

$u=x^4$ gives $du=4x^3\,dx$, so $x^3\,dx=\tfrac14\,du$ — the numerator supplies exactly the derivative needed: …

Common Mistakes

Mistake 1: Not seeing $x^8=(x^4)^2$.

Why it's wrong: missing this hides the arcsine structure. Correct approach: rewrite the power under the root as a perfect square.

Mistake 2: Dropping the $\tfrac14$ factor.

Why it's wrong: $du=4x^3\,dx$, so $x^3\,dx=\frac14\,du$; forgetting the $\frac14$ scales the answer. Correct approach: solve $du$ for $x^3\,dx$. …