Evaluate the following integrals:

Assume $x^{4}+1=t$

$d\left(x^{4}+1\right)=d t$

$4 x^{3} d x=d t$

$x^{3} d x=\frac{d t}{4}$

Substituting $\mathrm{t}$ and $\mathrm{dt}$

$\Rightarrow \int \frac{1}{4} \sin t d t$

$\Rightarrow \frac{-1 \cos t}{4}+c$

But $t=x^{4}+1$

$\Rightarrow \frac{-1}{4} \cos \left(x^{4}+1\right)+c$