Please Help integrating $P(t) = 3 + 2 \cos((1+t)/26)\pi$ I am given the equation:
$$P(t) = 3 + 2 \cos \left(\frac{1+t}{26}\right)\pi$$  
Where $t$ = time in weeks and $P$ = pollutant released into the atmosphere
I am told to integrate:
$$\int_0^4 3 + 2 \cos \left(\frac{1+t}{26} \right)\pi  dt$$
in stages:
(i) $$\int_0^4 3 dt$$
(ii) $$\int_0^4 2 \cos \left(\frac{1+t}{26}\right)\pi dt$$ Where, for (ii), I am told to use the substitution $u = \left(\frac{1+t}{26}\right)\pi$
I know that after integrating, (i) $= 12$.
For (ii), I have tried differentiating $u$ with respect to $t$ using the quotient rule with varying results. I'm not sure what I'm doing wrong, so any help would be appreciated. Thanks. 
 A: To evaluate 
$$\int_{0}^{4} 2 \cos \left( \frac{\pi(1+t)}{26} \right)  \mathrm{d}t$$
substitute $\frac{\pi(1+t)}{26} = u$ and apply $\sin A - \sin B$ identity to simplify your answer. 
A: Your substitution here is
\begin{equation}
u(t) = \frac{\pi(1+t)}{26}
\end{equation}
Notice the type of function that $u$ is linear. It's derivative is simply
\begin{equation}
\frac{du}{dt} = \frac{\pi}{26}.
\end{equation}
Then your integral becomes
\begin{equation}
\frac{1}{13} \int_{\frac{\pi}{26}}^{\frac{5 \pi}{26}} cos(u) du
\end{equation} 
A: Do you mean to find $$\int_0^4 2\cos\bigr(\textstyle{1+t\over 26}\bigl)\pi dt$$ 
or
 $$\int_0^4 2\cos\bigr(\textstyle{1+t\over 26}\pi\bigl) dt\ ?$$
 They are different things.
I assume it's the latter integral as you were asked to substitute 
$$\tag{1}u= {1+t\over26} \pi.$$
Now, to find $du$, you have to differentiate $ ({1+t\over26})\pi $.  You could use the quotient rule to do this:
$$ \textstyle 
du = \Bigl[ {(1+t)\pi\over26} \Bigr]'\,dt=  \Bigl[ {((1+t)\pi)'\cdot 26  -(1+t)\pi(26)'\over 26^2}\Bigr]\,dt
= \Bigl[{\pi\cdot 26-(1+t)\pi\cdot 0\over 26^2 }\Bigr]\,dt
={\pi\over 26}\,dt.
$$
But, this is more work than needed. Though you do have a quotient, the denominator and the $\pi$ upstairs in $(1)$ are constants and you can take advantage of the fact that you can factor multiplicative constants out when taking derivatives. The process of finding the derivative is  simpler to do then:
$$
 {1+t\over26} \pi = {\pi\over26}({1+t})
$$
and $\pi/26$ is just a number; so
$${du }=
 \Bigl[ {1+t\over26} \pi\Bigr]'\,dt
= \Bigl[{\pi\over26}({1+t})\Bigr]'\,dt= {\pi\over 26} (1+t)'\,dt={\pi\over 26}\cdot1\,dt={\pi\over26}\,dt.
$$ 
