Evaluate the integral: $\int \frac{x^3}{\sqrt{x^2 - 16}} \, dx$

Neetesh Kumar | February 5, 2025

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This is the solution to Integral Problem

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Step-by-step solution:

We need to evaluate the integral:

$\int \frac{x^3}{\sqrt{x^2 - 16}} \, dx$

Step 1: Use Trigonometric Substitution

Since we have a square root expression of the form $\sqrt{x^2 - a^2}$, we use the substitution:

$x = 4\sec \theta$

Differentiating both sides:

$dx = 4\sec \theta \tan \theta \, d\theta$

Also, using the identity $\sec^2 \theta - 1 = \tan^2 \theta$:

$\sqrt{x^2 - 16} = \sqrt{16\sec^2 \theta - 16} = \sqrt{16(\sec^2 \theta - 1)} = \sqrt{16\tan^2 \theta} = 4\tan \theta$

Step 2: Rewrite the Integral

Substituting $x = 4\sec \theta$ into the integral:

$\int \dfrac{(4\sec \theta)^3}{\sqrt{(4\sec \theta)^2 - 16}} \cdot 4\sec \theta \tan \theta \, d\theta$

Simplify the terms:

$\int \dfrac{64\sec^3 \theta}{4\tan \theta} \cdot 4\sec \theta \tan \theta \, d\theta$

Cancel out $\tan \theta$:

We get: $\int 64\sec^4 \theta \, d\theta$

Step 3: Use the Integral of $\sec^4 \theta$

We use the identity:

$\sec^4 \theta = (\sec^2 \theta)^2$

Using the standard integral formula:

$\int \sec^4 \theta \, d\theta = \frac{1}{3} \tan^3 \theta + \tan \theta + C$

Thus:

$\int 64\sec^4 \theta \, d\theta = 64 \left( \frac{1}{3} \tan^3 \theta + \tan \theta \right) + C$

Step 4: Substitute Back $\theta$

Since $x = 4\sec \theta$, we have:

$\tan \theta = \dfrac{\sqrt{x^2 - 16}}{4}$

Substituting back:

$\dfrac{64}{3} \left(\dfrac{\sqrt{x^2 - 16}}{4} \right)^3 + 64 \cdot \dfrac{\sqrt{x^2 - 16}}{4} + C$

Simplify:

$\dfrac{64}{3} \cdot \dfrac{(x^2 - 16)^{3/2}}{64} + 16\sqrt{x^2 - 16} + C$

After solving further:

$\dfrac{(x^2 - 16)^{\frac{3}{2}}}{3} + 16\sqrt{x^2 - 16} + C$

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Final Answer:

$\boxed{\dfrac{(x^2 - 16)^{\frac{3}{2}}}{3} + 16\sqrt{x^2 - 16} + C}$

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