<s>[INST] Evaluation of $\int_{0}^{1}\frac{x^4(1-x^2)^5}{(1+x^2)^{10}}dx$ 
Evaluation of $\displaystyle \int_{0}^{1}\frac{x^4(1-x^2)^5}{(1+x^2)^{10}}dx$

$\bf{My\; Try::}$ Let $$I = \int_{0}^{1}\frac{x^4(1-x^2)^5}{(1+x^2)^{10}}dx$$
Now Put $x=\tan \theta\;,$ Then $dx = \sec^2 \theta d\theta$ and changing limits, We get
$$I = \int_{0}^{\frac{\pi}{4}}\frac{\tan^4 \theta\cdot \left(1-\tan^2 \theta\right)^5}{(1+\tan^2\theta)^{10}}\cdot \sec^2 \theta d\theta = \int_{0}^{\frac{\pi}{4}}\sin^4 \theta\cdot \cos^4 \theta \left(\cos 2 \theta\right)^5 d\theta$$
So $$I = \frac{1}{16}\int_{0}^{\frac{\pi}{4}}(\sin 2\theta)^4\cdot (\cos 2 \theta)^4\cdot \cos 2 \theta d\theta$$
Now Put $\sin 2 \theta = t\;,$ Then $2\cos 2 \theta d\theta = dt$ and changing limits, We get
$$I = \frac{1}{64}\int_{0}^{1}t^4(1-t^2)^2 dt$$
Now after expanding , We can integrate it,
My question is can we solve it any other way, If yes then plz explain here
Thanks
 [/INST] 
There was a small error in the development in the OP.  The integral $I$ should be $$I=\frac{1}{32}\int_0^1 t^4(1-t^2)^2\,dt$$

We can evaluate the integral of interest in terms of the Beta function, $$B(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}\,dt$$and its relationship to the Gamma function, $$B (x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$
Note that 
$$\begin{align}
I&=\frac{1}{32}\int_0^1 t^4(1-t^2)^2\,dt\\\\
&=\frac{1}{64}\int_0^1 t^{3/2}(1-t)^2 \,dt\,\,\dots\text{enforcing the substitution}\,\,t\to \sqrt{t}\\\\
&=\frac{1}{64}B(5/2,3)\\\\
&=\frac{1}{64}\frac{\Gamma(5/2)\Gamma(3)}{\Gamma(11/2)}\\\\
&=\frac{1}{64}\frac{\left(3\frac{\sqrt{\pi}}{4}\right)\,(2!)}{945\frac{\sqrt{\pi}}{32}}\\\\
&=\frac{1}{1260}
\end{align}$$
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