Question

Simplify:

$\frac{x^3\sin \left({\tan }^{-1}{x}^4\right)}{1+x^8}$

Answer 1

$\frac{x^3\sin \left({\tan }^{-1}{x}^4\right)}{1+x^8}$

$\Rightarrow x^3\frac{x^4.\left(\sqrt{1+(x^4{)}^2}\right)}{(1+(x^4{)}^2)}\frac{1}{1+x^8}$

$\Rightarrow x^3.\frac{x^4\sqrt{1+(x^4{)}^2}}{(1+(x^4{)}^2)}\frac{1}{1+x^8}$

$=\frac{x^4\sqrt{1+x^8}}{(1+x^8{)}^2}.x^3$

$=\frac{x^7}{(1+x{)}^{3/2}}$