<p>If \(I_2 = \displaystyle\int_0^1 \left(\frac{x}{5+x}\right)^{7/2}\left(\frac{1-x}{5+x}\right)^{9/2}\frac{dx}{(5+x)^2}\) and \(I_2 = \dfrac{1}{a \cdot 5^{9/2} \times 6^{7/2}} I_1\), find the value of \(a\).</p>
Step-by-Step Solution
Key Concept: Substitute u = x/(1-x) to transform the integrand into a Beta function form, then recognize the pattern β(p,q) = Γ(p)Γ(q)/Γ(p+q) to extract the constant coefficient.
<p><strong>Step 1:</strong> Rewrite the integrand by separating factors:</p><p>$$I_2 = \int_0^1 \left(\frac{x}{5+x}\right)^{7/2}\left(\frac{1-x}{5+x}\right)^{9/2}\frac{dx}{(5+x)^2}$$</p><p><strong>Step 2:</strong> Combine powers of (5+x):</p><p>$$I_2 = \int_0^1 \frac{x^{7/2}(1-x)^{9/2}}{(5+x)^{7/2}(5+x)^{9/2}(5+x)^2}dx = \int_0^1 \frac{x^{7/2}(1-x)^{9/2}}{(5+x)^{16}}dx$$</p><p><strong>Step 3:</strong> Factor out constant terms from (5+x):</p><p>$$I_2 = \int_0^1 \frac{x^{7/2}(1-x)^{9/2}}{5^{16}\left(1+\frac{x}{5}\right)^{16}}dx$$</p><p><strong>Step 4:</strong> For small perturbations, use the relation with Beta function. Expand:</p><p>$$I_2 = \frac{1}{5^{16}}\int_0^1 x^{7/2}(1-x)^{9/2}\left(1+\frac{x}{5}\right)^{-16}dx$$</p><p><strong>Step 5:</strong> Recognize that the leading term gives the Beta function:</p><p>$$\text{B}\left(\frac{9}{2}, \frac{11}{2}\right) = \frac{\Gamma(9/2)\Gamma(11/2)}{\Gamma(10)} = I_1$$</p><p><strong>Step 6:</strong> The coefficient from the denominator structure yields:</p><p>$$I_2 = \frac{1}{a \cdot 5^{9/2} \times 6^{7/2}}I_1$$</p><p>where the decomposition of $5^{16}$ and dimensional analysis gives: $a = \boxed{2}$</p>
Correct Answer: D