Question

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

Answer 1

Let $ta{n}^{-1}{x}^4=t$

Differentiating both side of above equation with respect to $x$.

$\frac{d\left({\tan }^{-1}{x}^4\right)}{dx}=\frac{dt}{dx}$

$\frac{4x^3}{1+x^8}=\frac{dt}{dx}$ $[{\tan }^{-1}x=\frac{1}{1+x^2}]$

$\frac{x^3}{1+x^8}dx=\frac{1}{4}dt$

Substitute ${\tan }^{-1}{x}^4=t$ and $\frac{x^3}{1+x^8}dx=\frac{1}{4}dt$ in equation (1).

$I=\frac{1}{4}\int \sin \left(t\right)dt$

$=\frac{1}{4}(-\cos t)+C$ (2)

Substitute $t={\tan }^{-1}{x}^4$ in equation(2)

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

Thus, $\int \left(\frac{(x^3\sin \left(ta{n}^{-1}{x}^4\right)}{1+x^4}\right)dx$ is equal to $\frac{-\cos \left({\tan }^{-1}{x}^4\right)}{4}+C$ where $C$ is integration constant.