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I'm given a PDE $u_{tt}=u_{xx}+t\cos(x)$ with BC $u_x(t,0)=-1$, $u_x(t,\pi)=0$ and IC $u(0,x)=\cos(x)$, $u_t(0,x)=\cos^2(x)$, and I'm told to compute $$\int_0^\pi u(t,x) dx$$

I've already taken the integral of the whole equation, but now I'm left with $$\pi u_{tt}=1$$

In order to fully compute the quantity, I figured I should integrate it once with respect to $t$ so I can utilize the IC $u_t(0,x)=\cos^2(x)$. But I'm uncertain over which interval to evaluate this. Would it be $\int_0^\pi dt$?

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    Sometimes it is helpful to multiply the equation by $u$, $u_t$ or $u_x$ and integrate using also integration by parts. – 2017-02-14
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    Agree with @JohnB, integration by parts (or more importantly the divergence theorem) is always the second, if no the first thing to try when one deals with PDEs (ODEs too!). – 2017-02-14

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Integrating the equation you obtain $$ \int_0^\pi u_{tt}(t,x)\,dx=\frac{1}{\pi}. $$ Assuming that we can differentiate under the integral sign we have $$ \frac{d^2}{dt^2}\int_0^\pi u(t,x)\,dx=\int_0^\pi u_{tt}(t,x)\,dx=\frac{1}{\pi}\implies \int_0^\pi u(t,x)\,dx=\frac{t^2}{2\,\pi}+A\,t+B $$ for some constants $A,B$. You can determine their values from the initial conditions.