Edwards Penney, Calculus: Early Transcendentals, 7th Ed., Ch 5.6 Ex 36: how is the integral in the solutions manual calculated? In the Calculus book "Calculus: Early Transcendentals" by Edwards and Penney, 7th Edition, in the chapter on the fundamental theorem of calculus, there is an exercise (Chapter 5.6, exercise 36) about the average value of a continuous function on a closed interval. The function is $$T(t)=80+10 \sin \left(\frac{\pi}{12}(t-10)\right)$$ and the average value is calculated on the interval [12, 18].
For me it was straightforward to calculate the average: $$\overline{T}=\frac{1}{18-12} \int_{12}^{18}T(t)dt$$
But my question is specifically about the calculation of the indefinite integral $$\int T(t)dt$$
My calculation is:
$$80t + 10\left(-\frac{12}{\pi}\cos {\frac{\pi}{12}(t-10)}\right)$$
The solution uses a different calculation, which I do not understand:
$$ \int T(t)dt = \frac{40}{\pi} \left( 2\pi(t-10)-3\cos {\frac{\pi(t-10)}{12}} \right) $$
My question is how did they calculate the antiderivative in the line above? Where does the $2\pi(t-10)$ com from for example?
Note that maple gives
$$ \int T(t)dt = 80t + \frac{120 \cos {\frac{1}{12}\pi t + \frac{1}{6} \pi}}{\pi}$$
which is also different, though when I graph $$\cos {\frac{1}{12}\pi t + \frac{1}{6} \pi}$$ I can see that it is the negative of $$\cos {\frac{\pi}{12}(t-10)},$$
so my calculation is the same as that of maple actually.
 A: In comments, you mention you want to know how the manual reached the solution they did. So here's what is most likely what happened.
You (possibly silently or in your head), essentially did the following:
$$\begin{align*}
\int \left(80+10 \sin \left(\frac{\pi}{12}(t-10)\right)\right)\,dt &= \int 80\,dt + \int 10\sin\left( \frac{\pi}{12}(t-10)\right)\,dt\\
&= 80t + 10\int\sin\left(\frac{\pi}{12}(t-10)\right)\,dt.
\end{align*}$$
You then make a change of variable, $u=\frac{\pi}{12}(t-10)$, $du = \frac{\pi}{12}\,dt$, so
$$\begin{align*}
\int\sin\left(\frac{\pi}{12}(t-10)\right)\,dt &= \int\frac{12}{\pi}\sin(u)\,du\\
&= -\frac{12}{\pi}\cos(u) + C\\
&= -\frac{12}{\pi}\cos\left(\frac{\pi}{12}(t-10)\right)+C
\end{align*}$$
and putting it all together, you picked the antiderivative
$$F(t) = 80t +10\left(-\frac{12}{\pi}\cos\left(\frac{\pi}{12}(t-10)\right)\right).$$
By contrast, they did the same change of variable, but they did it from the get-go:
$$\begin{align*}
\int \left(80+10 \sin \left(\frac{\pi}{12}(t-10)\right)\right)\,dt &=
\int\left( 80 + 10\left(\frac{12}{\pi}\sin(u)\right)\right)\,du\\
&= 80u + 10\left(-\frac{12}{\pi}\cos(u)\right) + C\\
&= 80\left(\frac{\pi}{12}(t-10)\right) \\
&\qquad\quad+ 10\left(-\frac{12}{\pi}\cos\left(\frac{\pi}{12}(t-10)\right)\right) + C.
\end{align*}$$
Then they factor out a $\frac{40}{\pi}$ (note that $\gcd(80,120) = 40$; that's why they factor it out; and the $\pi$ denominator is to have that irrational denominator out in front), and choose the constant $C=0$ to get
$$\begin{align*}
80\left(\frac{\pi}{12}(t-10)\right) &+ 10\left(-\frac{12}{\pi}\cos\left(\frac{\pi}{12}(t-10)\right)\right) \\
&= 
\frac{40}{\pi}\left( 2\pi(t-10) - 3\cos\left(\frac{\pi}{12}(t-10)\right)\right).
\end{align*}$$
Just between the two of us, yours is less complicated, more likely if you are doing it by hand.
