An Annoying Rational Function

The Weierstrass substitution of $x=\tan\left(\dfrac{\theta}{2}\right)$ yields $\sin\theta=\dfrac{2x}{1+x^2},$ $\cos\theta=\dfrac{1-x^2}{1+x^2},$ and $d\theta=\dfrac{2}{1+x^2}\ dx.$ This substitution is often used to simplify integrals with rational functions containing trigonometric functions, such as the integral of $\dfrac{3}{1+\cos x+2\sin x}.$

We can rewrite the integral as follows.

$\int_0^1\frac{x^4(1-x^2)^5}{(1+x^2)^{10}}\ dx=\frac{1}{32}\int_0^1\left(\frac{2x}{1+x^2}\right)^4\times\left(\frac{1-x^2}{1+x^2}\right)^5\times\frac{2}{1+x^2}\ dx$

Applying a reverse Weierstrass substitution, this is equivalent to $\dfrac{1}{32}\displaystyle\int_0^\frac{\pi}{2}\sin^4\theta\cos^5\theta\ d\theta.$ Using the identity $\sin^2\theta+\cos^2\theta=1,$ we can reduce this integral to a simple $u$ -substitution.

$\frac{1}{32}\int_0^\frac{\pi}{2}\sin^4\theta\cos^5\theta\ d\theta=\frac{1}{32}\int_0^\frac{\pi}{2}\sin^4\theta(1-\sin^2\theta)^2\cos\theta\ d\theta$

This lends itself to $u=\sin\theta$ and $du=\cos\theta\ d\theta.$

$\frac{1}{32}\int_0^\frac{\pi}{2}\sin^4\theta(1-\sin^2\theta)^2\cos\theta\ d\theta=\frac{1}{32}\int_0^1u^4(1-u^2)^2\ du$

Finally, we evaluate the integral of a polynomial.

$\frac{1}{32}\int_0^1u^4(1-u^2)^2\ du=\frac{1}{32}\int_0^1u^8-2u^6+u^4\ du=\frac{1}{32}\left(\frac{1}{9}-\frac{2}{7}+\frac{1}{5}\right)=\dfrac{1}{1260}$

Thus, the answer is $\boxed{1260}.$

Let the integral be $I$ , then we have:

$\begin{aligned} I & = \int_0^1 \frac{x^4(1-x^2)^5}{(1+x^2)^{10}} dx & \small \blue{\text{Let }x = \tan \theta, \space d x = \sec^2 \theta \space d\theta} \\ & = \int_0^\frac{\pi}{4} \frac{\frac{\sin^4 \theta}{\cos^4 \theta}\left(1-\frac{\sin^2 \theta}{\cos^2 \theta}\right)^5\sec^2 \theta}{\sec^{20} \theta} d \theta \\ & = \int_0^\frac{\pi}{4} \sin^4 \theta \cos^4 \theta \cos^5 (2\theta) \space d \theta \\ & = \frac{1}{16} \int_0^\frac{\pi}{4} \sin^4 (2\theta) \cos^5 (2\theta) \space d \theta \\ & = \frac 1{\blue{32}} \int_0^{\{blue{\frac \pi 2}} \sin^\red 4 \blue{(2\theta)} \cos^{\red 5} \blue{(2\theta)} \ d \blue{(2\theta)} & \small \blue{\text{Integrate with respect to } (2\theta)} \\ & = \frac 1{\red{64}} \red B \left(\frac{\red 4+1}2, \frac{\red 5+1}2 \right) & \small \red{\text{where } B(m,n) \text{ is Beta function}} \\ & = \frac{B \left(\frac{5}{2}, 3 \right)}{64} = \frac{\Gamma \left(\frac{5}{2}\right)\Gamma \left(3 \right)}{64 \Gamma \left(\frac{11}{2}\right)} & \small \blue{\Gamma(n) \text{ is Gamma function}} \\ & = \frac{\frac 32 \cdot \frac 12 \sqrt \pi \cdot 2!}{64\cdot \frac 92 \cdot \frac 72 \cdot \frac 52 \cdot \frac 32 \cdot \frac 12 \sqrt\pi} \\ & = \frac{1}{\boxed{1260}} \end{aligned}$

Not really a solution, but this was one fire integral Trevor!