Ratio Of Integrals

The substitution $x \; = \; \frac{5\sin^2\theta}{5+\cos^2\theta}$ gives $\frac{x}{5+x} \; = \; \tfrac16\sin^2\theta \hspace{2cm} \frac{1-x}{5+x} \; = \; \tfrac15\cos^2\theta \hspace{2cm} \frac{1}{(5+x)^2}\,dx \; = \; \frac{1}{15}\sin\theta\cos\theta\,d\theta$ and hence $\begin{aligned} I_2 & = \; \frac{1}{15}\int_0^{\frac12\pi} \left(\tfrac16\sin^2\theta\right)^{\frac72}\left(\tfrac15\cos^2\theta\right)^{\frac92}\,\sin\theta\cos\theta\,d\theta \; = \; \frac{1}{15 \times 6^{\frac72} \times 5^{\frac92}} \int_0^{\frac12\pi} \sin^8\theta\cos^{10}\theta\,d\theta \\ & = \; \frac{1}{30 \times 6^{\frac72} \times 5^{\frac92}} B(\tfrac92,\tfrac{11}{2}) \; = \; \frac{1}{6^{\frac72} \times 5^{\frac92}}I_1 \end{aligned}$ For those who don't like this substitution, we can instead calculate $\begin{aligned} I_2 & = \; \int_0^1 \frac{x^{\frac72}(1-x)^{\frac92}}{(5+x)^{10}}\,dx \; = \; \frac{1}{5^{10}}\int_0^1 x^{\frac72}(1-x)^{\frac92} \big(1 + \tfrac{x}{5}\big)^{-10}\,dx\\ & = \; \frac{1}{5^{10}}\sum_{n \ge 0} \frac{(-1)^n}{5^n} \binom{n+9}{n}\int_0^1 x^{n+\frac72}(1-x)^{\frac92}\,dx \; = \; \frac{1}{5^{10}}\sum_{n \ge 0}\frac{(-1)^n}{5^n}\binom{n+9}{n} B\big(n + \tfrac92,\tfrac{11}{2}\big) \\ &= \; \frac{1}{5^{10}}\sum_{n \ge 0}\frac{(-1)^n}{5^n}\binom{n+9}{n} \frac{(\frac72+n)\times(\frac52+n)\times(\frac32+n) \cdots\times \frac{11}{2}\times\frac92}{(n+9)\times(n+8)\times(n+7)\times \cdots\times 11 \times10}B\big(\tfrac92,\tfrac{11}{2}\big) \\ & = \; \frac{1}{5^{10}}\sum_{n\ge 0}\frac{(-1)^n}{5^n} \frac{(\frac72+n)\times(\frac52+n)\times(\frac32+n) \cdots\times \frac{11}{2}\times\frac92}{n!}B\big(\tfrac92,\tfrac{11}{2}\big) \\ & = \; \frac{1}{5^{10}}\sum_{n \ge 0} \frac{(-\frac92)(-\frac{11}{2})\cdots(-\frac52-n)(-\frac72-n)}{n! 5^n}B\big(\tfrac92,\tfrac{11}{2}\big) \; = \; \frac{1}{5^{10}}\left(1 + \frac{1}{5}\right)^{-\frac92}B\big(\tfrac92,\tfrac{11}{2}\big) \\ & = \; \frac{1}{5^{\frac{11}{2}} \times 6^{\frac{9}{2}}}B\big(\tfrac92,\tfrac{11}{2}\big) \; = \; \frac{1}{5^{\frac92} \times 6^{\frac72}}I_1 \end{aligned}$ Either way we deduce that $\frac{I_1}{I_2} \; = \; 5^{\frac92} \times 6^{\frac72} \; = \; 5 \times 30^{\frac72}$ making the answer $\boxed{30}$ .

Rewrite the Integral as $\large I_2=\int_{0}^{1}\left(\dfrac{x}{5+x}\right)^{\frac 72} \left(\dfrac{1-x}{5+x}\right)^{\frac 92}\dfrac{dx}{(5+x)^2}$

and do the substitution $\dfrac{x}{5+x}=t$ so that $\dfrac{dx}{(5+x)^2}=\dfrac{dt}{5}$ and the integral becomes

$\large \dfrac{1}{(5)^{\frac {11}{2}}} \int_{0}^{\frac 16} (t)^{\frac 72} (1-6t)^{\frac 92} dt$ and now from here do the substitution $6t=u$ and we simply obtain $I_2=\dfrac{1}{5^{\frac 92} \times 6^{\frac 72}} I_1$ and we conclude $a=\boxed{30}$ .

I am feeling happy that my assymptotic method worked well. I found a very near value of $a$ in this assymptotic method which was $32.24$ . How? Let's look.

The curve of the function $f(x)=\frac{{x^{7/2}}{(1-x)^{9/2}}}{(x+5)^{10}}$ is just like the previous function $g(x)$ ,

$g(x)=x^{7/2}(1-x)^{9/2}$ , both having nodes at $0$ and $1$ only.

Now, we shall take the weighted average value of the denominator of $f(x)$ ,( $(x+5)^{10}$ ).

So, the weighted average is ...

$\frac{\int_{0}^1 (x+5)^{10} dx}{1}$

$=\frac{6^{11}-5^{11}}{11}$ .

So, we get $\frac{I_{1}}{I_{2}}=\frac{6^{11}-5^{11}}{11×30}$ . And comparing this with given form we get $a≈(\frac{6^{11}-5^{11}}{5×11×30})^{2/7}=32.24....$

From here we assume $a=30$ , being motivated by the bounds of $(x+5)$ which are $5$ and $6$ . And near to $32$ , $30$ has both $5$ and $6$ as it's divisors.