Not fully understanding the cosine angle addition identity in a definite integral problem

Question

For homework in my calculus class, I'm trying to show via u substitution that the following definite integral is equal to zero:

$$ \int_{0}^{2} (1-t) \cos(\pi t) \ dt $$

Here are the u substitution parameters I used:

• u = 1 - t
• du = -dt
• t = 1 - u
• When u is 0, t = 1
• When u is 2, t = -1

This is where I got stuck:

$$ -\int_{1}^{-1} u \ cos(\pi(1-u)) \ du $$

I checked the solution in the back of the book, and they had nearly the same intermediate equation (theirs lacks the leading negative sign). However, the next step confused me. I don't understand how they got from their intermediate equation to the following integral:

$$ \int_{1}^{-1} u [\cos(\pi)\cos(u) - \sin(\pi)\sin(u)] \ du $$

I expected this:

$$ \int_{1}^{-1} u [\cos(\pi)\cos(\pi u) + \sin(\pi)\sin(\pi u)] \ du $$

What step(s) am I missing in applying the cosine angle addition formula that allowed the textbook authors to arrive at their version of the integral?

Textbook: OpenStax Calculus Volume 1

Section: 5.5

Exercise: 311, Page 594

Answer 1

Looks like the answer is that the book is incorrect. Thanks to all who responded.

Answer 2

Here's a way by integration by parts for fun. Let $$I=\int_0^2\left(1-t\right)\cos(\pi t)dt=\int_0^2\cos(\pi t)dt-\int_0^2t\cos(\pi t)dt=\left(\frac{1}{\pi}\sin(\pi t)\bigg|^2_0\right)-\int_0^2t\cos(\pi t)dt$$

The last integral can be done by integration by parts by letting $$s=t\implies ds=dt$$ $$dr=\cos(\pi t)\space dt\implies r=\frac{1}{\pi}\sin(\pi t)$$

Hence $$-\int_0^2t\cos(\pi t)dt=-\frac{t}{\pi}\sin(\pi t)\bigg|^2_0+\frac{1}{\pi}\int_0^2\sin(\pi t)dt=\left(-\frac{2}{\pi}\sin(2\pi)+0\right)-\frac{1}{\pi^2}\cos(\pi t)\bigg|^2_0$$ $$=(0)-\frac{1}{\pi^2}\cos(2\pi)+\frac{1}{\pi^2}\cos(0)=0$$

In conclusion, $$I=\left(\frac{1}{\pi}\sin(\pi t)\bigg|^2_0\right)-0=\frac{1}{\pi}\sin(2\pi)+\frac{1}{\pi}\sin(0)=0$$

Here's another way via symmetry:

Let $$s=t-1$$

so that $ds=dt$. One could just as well enforce the substitution $s=1-t$ which would again lead to a symmetric integral of an odd function. So, we have: $$I=\int_{t=0}^{t=2}(1-t)\cos(\pi t)dt=-\int_{-1}^1s\cos(\pi(s+1))ds=-\int_{-1}^1s\cos(\pi s+\pi)ds$$ $$=-\int_{-1}^1\left(s\cos(\pi s)\cos(\pi)-s\sin(\pi s) \sin(\pi)\right)ds=\int_{-1}^1s\cos(\pi s)ds$$

The function $f(s):=s\cos(\pi s)$ is odd since $f(-s)=-s\cos(-\pi s)=-f(s)$ therefore, $$I=\int_{-1}^1s\cos(\pi s)ds=0$$

Answer 3

A slightly more direct route to coreyman's symmetry argument:

Make instead the substitution $u=2-t$. After cleaning up the signs, the original integral $I = \int_0^2 (1-t)\cos(\pi t)\,dt$ becomes $-\int_0^2 (1-u)\cos(2\pi - \pi u)\,du$. But $\cos$ is $2\pi$-periodic and even, so $\cos(2\pi - \pi u) = \cos(-\pi u) = \cos(\pi u)$. Hence $I = -\int_0^2 (1-u)\cos(\pi u)\,du = -I$ and so we must have $I=0$.