Class 11
Derivatives, derivatives, derivatives!
Derivatives!
Chain rule, product rule!
Do not mix variables in substitution!
Always update the limits!
Do not drop \(dx\)!
Do not force a substitution! If it does not fit, it does not work!
\(\displaystyle \color{red} \int f(x)g(x)\;dx \neq \int f(x)\;dx \int g(x)\;dx\)!!!!!
Do not drop the \(\displaystyle\lim_{x\to a}\)!
Especially when you do a substitution and the \(a\) changes!
\(\infty\cdot 0 \neq 0\)! Sometimes it is, but not always!
\(0^0 \neq 1\)! Sometimes it is, but not always!
Do not try to “calculate” with infinities! \(\infty\) and \(-\infty\) are not numbers, and the rules for working with them are complicated. It is easy to make a mistake. There is a reason why limits are used.
If it is not type \(\frac{0}{0}\) or type \(\frac{\infty}{\infty}\), do not try to use l’Hospital’s rule!
You can re-do any problem on which you did not get full credit.
The corrections are due next Thursday, 2/29, in class.
Turn in the corrections and the original exam!
Do not write the corrections on the original exam! Use separate sheets of paper. Write down the problem number clearly, so I can see which problem you are doing.
The solutions must be written in clear, organized, and mathematically correct way. Use proper notation and terminology. It is a take-home assignment, so you have time to write everything down carefully.
I will grade the corrections, and average the new score with the old one. The resulting score will be capped at 100.
\[\begin{align*} \int \sec^3 x\;dx &\class{fragment}{{}= \int\sec^2 x\sec x\;dx}\\ & \class{fragment}{\color{blue}\qquad u = \sec x\qquad\qquad\qquad dv = \sec^2x\;dx}\\ & \class{fragment}{\color{blue}\qquad du = \sec x \tan x\;dx\qquad v = \tan x}\\ &\class{fragment}{{}=\sec x\tan x - \int\tan^2 x \sec x\;dx}\class{fragment}{{}= \sec x\tan x - \int\left(\sec^2 x - 1\right)\sec x\;dx}\\[1.2em] &\class{fragment}{{}=\sec x\tan x + \int\sec x\;dx - \int\sec^3 x\;dx}\\[1.2em] &\class{fragment}{{}= \sec x\tan x + \ln\left\lvert \sec x + \tan x\right\rvert - \int \sec^3 x\;dx} \end{align*}\]
\(\displaystyle \int_0^{\pi/2} \cos(3x)\sin(5x)\;dx\)
Product to sum formulas:
\[\begin{align*} \sin u \cos v &= \frac{1}{2}\left[ \sin(u+v) + \sin(u-v) \right]\\ \cos u \sin v &= \frac{1}{2}\left[ \sin(u+v) - \sin(u-v) \right]\\ \cos u \cos v &= \frac{1}{2}\left[ \cos(u+v) + \cos(u-v) \right]\\ \sin u \sin v &= \frac{1}{2}\left[ \cos(u-v) - \cos(u+v) \right]\\ \end{align*}\]
\(\displaystyle \int_{-\pi}^{\pi} \cos(mx)\sin(nx)\;dx\), \(m\) and \(n\) non-negative integers.
\(\cos u \sin v = \frac{1}{2}\left[ \sin(u+v) - \sin(u-v) \right]\)
\(\displaystyle \int_{-\pi}^{\pi} \cos(mx)\cos(nx)\;dx\), \(m\) and \(n\) non-negative integers.
\(\cos u \cos v = \frac{1}{2}\left[ \cos(u+v) + \cos(u-v) \right]\)
\(\displaystyle \int_0^{1/2} \sqrt{1 - x^2}\;dx\)
\(\displaystyle \int \frac{\sqrt{9 - x^2}}{x}\;dx\)
\[\begin{align*} 1 - \cos^2 t &= \sin^2 t\\ 1 - \sin^2 t &= \cos^2 t\\ 1 + \tan^2 t &= \sec^2 t\\ \sec^2 t - 1 &= \tan^2 t \end{align*}\]
\(\displaystyle \int \sqrt{4 + x^2}\;dx\)
\(\displaystyle \int \frac{1}{\sqrt{x^2 - 16}}\;dx\)