Answer
$a.\quad \displaystyle \frac{15}{16}$
$ b.\quad$ 0
Work Step by Step
$a.$
Applying th.7, let $u=g(t)=1+t^{4}$ (a continuous function).
Then, $\displaystyle \quad du=g'(t)dt=4t^{3}dt,\qquad t^{3}dt=\frac{du}{4}$
and, the borders change to $g(0)=1$ and $g(1)=2$.
$\displaystyle \int_{0}^{1}t^{3}(1+t^{4})^{3}dt=\int_{1}^{2}u^{3}\cdot\frac{1}{4}du$
$=\displaystyle \frac{1}{4}\left[ \frac{u^{3+1}}{3+1} \right]_{1}^{2}$
$=\displaystyle \frac{1}{4}\left[ \frac{u^{4}}{4} \right]_{1}^{2}$
$=\displaystyle \frac{1}{16}(2^{4}-1^{4})$
$=\displaystyle \frac{15}{16}$
$b.$
Applying th.7, let $u=g(t)=1+t^{4}$ (a continuous function).
Then, $\displaystyle \quad du=g'(t)dt=4t^{3}dt,\qquad t^{3}dt=\frac{du}{4}$
and, the borders change to $g(-1)=2$ and $g(1)=2$.
$\displaystyle \int_{-1}^{1}t^{3}(1+t^{4})^{3}dt=\int_{2}^{2}u^{3}\cdot\frac{1}{4}du$
$=0$
(zero width integral)
