Question

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

Answer 1

Let x4 = t

∴ 4x3 dx = dt

$\Rightarrow \int\frac{x^3 \sin(\tan^{-1}x^4)}{1+x^8}dx=\frac{1}{4}\int\frac{\sin (\tan^{-1})}{1+t^2}dt$

Let $\tan^{-1}t=u$

∴ $\frac{1}{1+t^2}dt = du$

From (1), we obtain

$\int\frac{ x^3\sin(tan^{-1}x^4)dx}{1+x^8}=\frac{1}{4}\int\sin u \; du$

= $\frac{1}{4}(- \cos u)+C$

=$\frac{-1}{4}\cos(\tan^{-1}t)+C$

=$\frac{-1}{4}\cos (\tan^{-1}x^4)+C$