Question:
The value of the integral
\[
\int_{0}^{2} x^8 \left( \frac{4}{x^2} - 1 \right)^{\frac{5}{2}} \, dx
\]
is equal to:
The value of the integral
\[
\int_{0}^{2} x^8 \left( \frac{4}{x^2} - 1 \right)^{\frac{5}{2}} \, dx
\]
is equal to:
Show Hint
When the integrand contains a power of $x$ outside a parenthesis with $1/x^2$, always try to distribute a power of $x$ inside the parenthesis first to simplify the radical before choosing a substitution.
Updated On: May 20, 2026
- \( \frac{2^{15}}{63} \)
- \( \frac{2^{16}}{315} \)
- \( \frac{2^{16}}{189} \)
- \( \frac{2^{10}}{63} \)
Show Solution
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The Correct Option is D
Solution and Explanation
Concept:
To solve this integral, we first simplify the algebraic expression within the parenthesis. This often reveals a structure suitable for trigonometric substitution, specifically involving the form \( \sqrt{a^2 - x^2} \).
• Algebraic Simplification: \( \left( \frac{4 - x^2}{x^2} \right)^{n} = \frac{(4 - x^2)^n}{x^{2n}} \).
• Trigonometric Substitution: Use \( x = 2\sin \theta \) for terms involving \( \sqrt{4 - x^2} \).
• Wallis' Formula: A shortcut for integrals of the form \( \int_{0}^{\pi/2} \sin^m \theta \cos^n \theta \, d\theta \).
Step 1: Simplifying the integrand.
Rewrite the expression inside the integral: \[ x^8 \left( \frac{4 - x^2}{x^2} \right)^{5/2} = x^8 \frac{(4 - x^2)^{5/2}}{(x^2)^{5/2}} = x^8 \frac{(4 - x^2)^{5/2}}{x^5} = x^3 (4 - x^2)^{5/2} \] So the integral becomes: \[ I = \int_{0}^{2} x^3 (4 - x^2)^{5/2} \, dx \]
Step 2: Applying trigonometric substitution.
Let \( x = 2\sin \theta \), which implies \( dx = 2\cos \theta \, d\theta \).
• When \( x = 0, \theta = 0 \).
• When \( x = 2, \theta = \pi/2 \). The term \( (4 - x^2)^{5/2} = (4\cos^2 \theta)^{5/2} = 2^5 \cos^5 \theta \). Substituting everything into the integral: \[ I = \int_{0}^{\pi/2} (2\sin \theta)^3 (2^5 \cos^5 \theta) (2\cos \theta) \, d\theta \] \[ I = 2^3 \times 2^5 \times 2 \int_{0}^{\pi/2} \sin^3 \theta \cos^6 \theta \, d\theta = 2^9 \int_{0}^{\pi/2} \sin^3 \theta \cos^6 \theta \, d\theta \]
Step 3: Evaluating using Wallis' Formula.
For \( m=3 \) (odd) and \( n=6 \) (even), the constant \( K = 1 \): \[ \int_{0}^{\pi/2} \sin^3 \theta \cos^6 \theta \, d\theta = \frac{(2) \cdot (5 \cdot 3 \cdot 1)}{9 \cdot 7 \cdot 5 \cdot 3 \cdot 1} = \frac{2}{9 \cdot 7} = \frac{2}{63} \] Now, calculate the final value: \[ I = 2^9 \times \frac{2}{63} = \frac{2^{10}}{63} \]
• Algebraic Simplification: \( \left( \frac{4 - x^2}{x^2} \right)^{n} = \frac{(4 - x^2)^n}{x^{2n}} \).
• Trigonometric Substitution: Use \( x = 2\sin \theta \) for terms involving \( \sqrt{4 - x^2} \).
• Wallis' Formula: A shortcut for integrals of the form \( \int_{0}^{\pi/2} \sin^m \theta \cos^n \theta \, d\theta \).
Step 1: Simplifying the integrand.
Rewrite the expression inside the integral: \[ x^8 \left( \frac{4 - x^2}{x^2} \right)^{5/2} = x^8 \frac{(4 - x^2)^{5/2}}{(x^2)^{5/2}} = x^8 \frac{(4 - x^2)^{5/2}}{x^5} = x^3 (4 - x^2)^{5/2} \] So the integral becomes: \[ I = \int_{0}^{2} x^3 (4 - x^2)^{5/2} \, dx \]
Step 2: Applying trigonometric substitution.
Let \( x = 2\sin \theta \), which implies \( dx = 2\cos \theta \, d\theta \).
• When \( x = 0, \theta = 0 \).
• When \( x = 2, \theta = \pi/2 \). The term \( (4 - x^2)^{5/2} = (4\cos^2 \theta)^{5/2} = 2^5 \cos^5 \theta \). Substituting everything into the integral: \[ I = \int_{0}^{\pi/2} (2\sin \theta)^3 (2^5 \cos^5 \theta) (2\cos \theta) \, d\theta \] \[ I = 2^3 \times 2^5 \times 2 \int_{0}^{\pi/2} \sin^3 \theta \cos^6 \theta \, d\theta = 2^9 \int_{0}^{\pi/2} \sin^3 \theta \cos^6 \theta \, d\theta \]
Step 3: Evaluating using Wallis' Formula.
For \( m=3 \) (odd) and \( n=6 \) (even), the constant \( K = 1 \): \[ \int_{0}^{\pi/2} \sin^3 \theta \cos^6 \theta \, d\theta = \frac{(2) \cdot (5 \cdot 3 \cdot 1)}{9 \cdot 7 \cdot 5 \cdot 3 \cdot 1} = \frac{2}{9 \cdot 7} = \frac{2}{63} \] Now, calculate the final value: \[ I = 2^9 \times \frac{2}{63} = \frac{2^{10}}{63} \]
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