Definite Trig Integrals: Changing Limits of Integration $$\int_0^{\pi/4} \sec^4 \theta \tan^4 \theta\; d\theta$$
I used the substitution: let $u = \tan \theta$ ... then $du = \sec^2 \theta \; d\theta$.
I know that now I have to change the limits of integration, but am stuck as to how I should proceed.
Should I sub the original limits into $\tan \theta$ or should I let $\tan \theta$ equal the original limits and then get the new limits?
And if it help, the answers of the definite integral is supposed to be $0$.
Thanks in advance.

\begin{align}
I &= \int_{0}^{\pi/4} \sec^{4}(\theta) \tan^{4}(\theta) \ d\theta \\
&= \int_{0}^{\pi/4} \sec^{2}(\theta) ( 1 + \tan^{2}(\theta)) \tan^{4}(\theta) \ d\theta \\
&= \int_{0}^{\pi/4} \tan^{4}(\theta) \ d(\tan\theta) + \int_{0}^{\pi/4} \tan^{6}(\theta) \ d(\tan\theta) \\
&= \left[ \frac{1}{5} \tan^{5}(\theta) \right]_{0}^{\pi/4} + \left[ \frac{1}{7} \tan^{7}(\theta) \right]_{0}^{\pi/4} \\
&= \frac{1}{5} + \frac{1}{7} = \frac{12}{35}.
\end{align}
