Evaluate the following integrals:

Assume $x^{4}=t$

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

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

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

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

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

$\Rightarrow \frac{-1 \text { cost }}{4}+c$

But $t=x^{4}$

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