Evaluate: $\int \dfrac{\sqrt{x}}{\sqrt{a^3-x^3}}\,dx$

Evaluate: ∫ dfrac√(x)√(a³-x³)dx

NCERT Exemplar · Q24

\int \dfrac{\sqrt{x}}{\sqrt{a^3-x^3}}\,dx

Yanam CbseShort· 3mImportance★★★★★

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Concept understanding — U Substitution

U Substitution: The Reverse Chain Rule

The chain rule differentiates composite functions: the derivative of $\sin(x^2)$ is $\cos(x^2) \cdot 2x$ — differentiate the outer function, then multiply by the derivative of the inside. Integration asks the reverse: given $\cos(x^2) \cdot 2x$ , find the original function. That's what u substitution does — it reverses the chain rule.

The Core Intuition

When an integral looks like "a function times the derivative of its inside," substitute the inside with $u$ and the derivative of the inside with $du$ . Consider:

\int 2x \cos(x^2) \, dx

Here $2x$ is the derivative of $x^2$ , and $x^2$ is the inside of $\cos(x^2)$ . Let $u = x^2$ , so $du = 2x \, dx$ :

\int \cos(u) \, du = \sin(u) + C = \sin(x^2) + C

Check: the derivative of $\sin(x^2)$ is $\cos(x^2) \cdot 2x$ .

The Precise Statement

\int f(g(x)) \cdot g'(x) \, dx = \int f(u) \, du \quad \text{where } u = g(x), \, du = g'(x) \, dx

Valid provided $g$ is differentiable and the resulting integral in $u$ is simpler.

The Step-by-Step Method

Identify a function $g(x)$ whose derivative $g'(x)$ also appears (possibly up to a constant factor).
Set $u = g(x)$ , compute $du = g'(x) \, dx$ .
Rewrite the entire integral in $u$ and $du$ — every $x$ and $dx$ must be replaced.
Integrate with respect to $u$ .
Substitute back $u = g(x)$ .

Watch out

You cannot mix variables. If any $x$ remains after substitution, you chose the wrong $u$ (or must solve for $x$ in terms of $u$ — rare).

A Second Example (with a constant factor)

Evaluate $\int x \sqrt{x^2 + 1} \, dx$ . Let $u = x^2 + 1$ , so $x \, dx = \frac{1}{2} du$ :

\int \sqrt{u} \cdot \tfrac{1}{2} du = \tfrac{1}{2} \cdot \tfrac{2}{3} u^{3/2} + C = \tfrac{1}{3} (x^2 + 1)^{3/2} + C

When Does It Work?

When the integrand is something times the derivative of something inside. Common patterns:

$x \cdot f(x^2)$ — derivative of $x^2$ is $2x$ , so $u = x^2$
$e^{g(x)} \cdot g'(x)$ — derivative of $g(x)$ appears
$\frac{g'(x)}{g(x)}$ — leads to $\log|g(x)|$ …

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The key is to rewrite the integrand so that the substitution $t = x^{3/2}$ (or equivalently $u = x^3/a^3$ ) reveals a standard $\arcsin$ form. The final result is $\frac{2}{3} \arcsin\!\left(\frac{x^{3/2}}{a^{3/2}}\right) + C$ .

When you see a square root of a difference like $\sqrt{a^3 - x^3}$ , your first thought should be: can I turn this into something like $\sqrt{1 - u^2}$ ? That’s the classic pattern for an inverse sine (or inverse cosine) integral. The problem is that the numerator is $\sqrt{x}$ , not a simple constant. So we need to find a substitution that absorbs the $\sqrt{x}$ into the differential.

Notice that the denominator has $x^3$ inside the square root. If we set $u = x^{3/2}$ , then $du = \frac{3}{2} x^{1/2}\,dx$ , which exactly contains the $\sqrt{x}\,dx$ from the numerator. That’s the insight: the $\sqrt{x}$ is not a nuisance — it’s the derivative of $x^{3/2}$ up to a constant factor.

Let’s work it through cleanly.

Rewrite the integral

I = \int \frac{\sqrt{x}}{\sqrt{a^3 - x^3}}\,dx

Choose the substitution Let $t = x^{3/2}$ . Then $t^2 = x^3$ , and

dt = \frac{3}{2} x^{1/2}\,dx \quad\Rightarrow\quad \sqrt{x}\,dx = \frac{2}{3}\,dt

Replace everything in terms of $t$ The denominator becomes $\sqrt{a^3 - t^2}$ . So

I = \int \frac{\frac{2}{3}\,dt}{\sqrt{a^3 - t^2}} = \frac{2}{3} \int \frac{dt}{\sqrt{a^3 - t^2}}

Recognise the standard form This is exactly $\displaystyle \int \frac{du}{\sqrt{A^2 - u^2}} = \arcsin\!\left(\frac{u}{A}\right) + C$ , with $A = a^{3/2}$ and $u = t$ . …

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Q.Evaluate: $\int \dfrac{\sqrt{x}}{\sqrt{a^3-x^3}}\,dx$

Concept understanding — U Substitution

U Substitution: The Reverse Chain Rule

The Core Intuition

The Precise Statement

The Step-by-Step Method

A Second Example (with a constant factor)

When Does It Work?

Method: Substituting $u=x^{3/2}$ to reach an arcsine form

Steps

Common Mistakes

Q.Evaluate: ∫xa3−x3 dx\int \dfrac{\sqrt{x}}{\sqrt{a^3-x^3}}\,dx∫a3−x3​x​​dx

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Q.Evaluate: $\int \dfrac{\sqrt{x}}{\sqrt{a^3-x^3}}\,dx$

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