Question

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

Answer 1

Let $x^4=t$
$\therefore 4x^{3}dx=dt$
$\Rightarrow \int \frac{x^3\sin \left({\tan }^{-1}{x}^4\right)}{1+x^8}dx=\frac{1}{4}\int \frac{sin\left(ta{n}^{-1}t\right)}{1+t^2}dt$ ......(1)
Let ${\tan }^{-1}t=u$
$\therefore \frac{1}{1+t^2}dt=du$
From (1), we obtain
$\int \frac{x^3\sin \left({\tan }^{-1}{x}^4\right)dx}{1+x^8}=\frac{1}{4}\int \sin udu$
$=\frac{1}{4}(-\cos u)+C$
$=\frac{-1}{4}\cos \left({\tan }^{-1}t\right)+C$
$=\frac{-1}{4}\cos \left({\tan }^{-1}{x}^4\right)+C$