Evaluate: $\int \dfrac{dx}{x\sqrt{x^4-1}}$ (Hint: Put $x^2=\sec\theta$)

Evaluate: ∫ dfracdxx√(x⁴-1) (Hint: Put x²=secθ)

NCERT Exemplar · Q26

\int \dfrac{dx}{x\sqrt{x^4-1}}

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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 substitution $x^2=\sec\theta$ turns $\sqrt{x^4-1}$ into $\tan\theta$ and collapses the whole integrand to a constant $\tfrac12$ , giving $\dfrac12\sec^{-1}(x^2)+C$ .

Why this substitution

Write $\sqrt{x^4-1}=\sqrt{(x^2)^2-1}$ . The identity $\sec^2\theta-1=\tan^2\theta$ is the perfect match for $(x^2)^2-1$ , so putting $x^2=\sec\theta$ removes the square root cleanly. The stray $x$ in the denominator combines with the $x$ coming from $dx$ to give $x^2$ , which then cancels against $\sec\theta$ .

Step 1 — Set up the substitution

Let $x^2=\sec\theta$ . Differentiating, $2x\,dx=\sec\theta\tan\theta\,d\theta$ , so

dx=\frac{\sec\theta\tan\theta}{2x}\,d\theta.

Step 2 — Simplify the radical

\sqrt{x^4-1}=\sqrt{(x^2)^2-1}=\sqrt{\sec^2\theta-1}=\sqrt{\tan^2\theta}=\tan\theta,

taking $\theta\in[0,\tfrac{\pi}{2})$ (the principal range where $\tan\theta\ge 0$ ), consistent with $\sec^{-1}$ .

Step 3 — Rewrite the integral

\int \frac{dx}{x\sqrt{x^4-1}} = \int \frac{1}{x\,\tan\theta}\cdot\frac{\sec\theta\tan\theta}{2x}\,d\theta = \int \frac{\sec\theta}{2x^2}\,d\theta.

The two $x$ factors produced $2x^2$ in the denominator, and $\tan\theta$ cancelled.

Step 4 — Use $x^2=\sec\theta$ again

Replacing $x^2$ by $\sec\theta$ , …

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Q.Evaluate: $\int \dfrac{dx}{x\sqrt{x^4-1}}$ (Hint: Put $x^2=\sec\theta$ )

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?

Why this substitution

Step 1 — Set up the substitution

Step 2 — Simplify the radical

Step 3 — Rewrite the integral

Step 4 — Use $x^2=\sec\theta$ again

Method: Substitution reducing $\int\frac{dx}{x\sqrt{x^{2n}-1}}$ to arcsecant

Steps

Common Mistakes

Q.Evaluate: ∫dxxx4−1\int \dfrac{dx}{x\sqrt{x^4-1}}∫xx4−1​dx​ (Hint: Put x2=sec⁡θx^2=\sec\thetax2=secθ)

Why this substitution

Step 1 — Set up the substitution

Step 2 — Simplify the radical

Step 3 — Rewrite the integral

Step 4 — Use x2=sec⁡θx^2=\sec\thetax2=secθ again

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Q.Evaluate: $\int \dfrac{dx}{x\sqrt{x^4-1}}$ (Hint: Put $x^2=\sec\theta$ )

Step 4 — Use $x^2=\sec\theta$ again

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