What’s this? A format innovation? Haven’t had one of those in years! I think it’s adequately explained by the title. Let’s go!

1 If we integrate $\int \frac{1}{1-x}\dx$, we get $-\ln |1-x|+C$. Where did the minus come from?

It comes from the chain rule, or integration by substitution, depending on your angle of attack. If you differentiate $\ln(1-x)$, you get $-\frac{1}{1-x}$, so you need to flip the sign to get $\frac{1}{1-x}$. Equally, if you let $u = 1-x$ to integrate, you need to replace $\dx$ with $\diff{x}{u} \d{u}$, which is $-\du$.

2 How does $\frac{\sqrt{x}}{x}$ become $\frac{1}{\sqrt{x}}$?

It’s because $x = \sqrt{x}\sqrt{x}$. If you replace the $x$ on the bottom accordingly, you can divide top and bottom by $\sqrt{x}$.

3 How does $\frac{2}{3}(x+1)^{3/2} - 2\sqrt{x+1}$ become $\frac{2}{3}(x-2)\sqrt{x+1}$?

This is because $\sqrt{x+1}$ is a common factor. If you take this out, you get $\sqrt{x+1}\left(\frac{2}{3}(x+1) - 2\right)$. Then you need to tidy up the bracket: if you turn it into a single fraction, you get $\sqrt{x+1}\left(\frac{2(x+1)-6}{3}\right)$, which is $\frac{2}{3}\sqrt{x+1}\left(x-2\right)$.

4 If I have an expansion for something like $\frac{1+x}{1+3x}$ and need to approximate $\frac{101}{103}$, how do I pick $x$?

I’d start by setting them equal to each other here: $\frac{1+x}{1+3x} = \frac{101}{103}$ becomes $103 + 103x = 101 + 303x$ and $200x = 2$ so $x=0.01$. Once you’ve done a few of them, you’ll start to spot the pattern.

5 Why does $\ln\left(\frac{\sqrt{2}}{2}\right) = -\frac{1}{2}\ln(2)$?

If you simplify the square root (as we did earlier on), you get $\ln\left(\frac{1}{\sqrt{2}}\right)$. That’s the same as $\ln\left(2^{-1/2}\right)$, which is $-\frac{1}{2}\ln(2)$.

6 If $\cot(x)=0$ and $0 < x \le \frac{\pi}{2}$, why does $x = \frac{\pi}{2}$ ?

$\cot(x) \equiv \frac{\cos(x)}{\sin(x)}$, so (since $\sin(x)$ is always finite), $\cos(x)$ must be 0. The only place that happens in that domain is $x =\frac{\pi}{2}$.

7 How do I find $k$ such that $2\bb{a} + k\bb{b}$ and $5\bb{a} + 3\bb{b}$ are parallel?

Two vectors are parallel if one is a multiple of the other. I’d probably multiply the first by 5 and the second by 2 to get $10\bb{a} + 5k\bb{b}$ and $10\bb{a} + 6\bb{b}$, from which it’s clear that $5k=6$ and $k = \frac{6}{5}$.

8 How would you integrate $\int x^2\sqrt{x-2}\dx$?

I’d ask “what’s ugly?” and try to make it less ugly. Here’ it’s the $x-2$, which I’d substitute away: let $u = x-2$, so $\diff{u}{x}=1$ and $x^2 = (u+2)^2$.

Now the integral is $\int (u+2)^2 u^{1/2}\du$, which I would expand (in two stages) and integrate directly: $\int (u^2 + 4u + 4)u^{1/2}\du$, or $\int u^{5/2} + 4u^{3/2} + 4u^{1/2}\du$, which is $\frac{2}{7}u^{7/2} + \frac{8}{5}u^{5/2} + \frac{8}{3}u^{3/2} + C$.

9 How would you integrate $\int\frac {2e^x}{e^{2x}-1}\dx$?

Let $u = e^x$ so that $\diff{u}{x} = e^x$. The integral becomes $\int \frac{2}{u^2 -1}\d{u}$. That falls to partial fractions as $\int \frac{1}{u-1} - \frac{1}{u+1}\du$, which integrates to $\ln|u-1| - \ln|u+1|+c$.

However, we introduced $u$ and need to replace it: final answer is $\ln|e^x - 1| - \ln(e^x+1) + c$. (The second log argument is always positive, so it doesn’t need the absolute value signs.)

10 How many solutions does $2 = \sin(x) + \sin^2(x) + \sin^3(x) + \dots$ have for $0 \le x \lt 2\pi$?

Oo, nice. A geometric sequence with $a=r=\sin(x)$ – which is (almost everywhere) small enough for an infite sum to exist.

When it does, that’s $S = \frac{a}{1-r} = \frac{\sin(x)}{1-\sin(x)}$, so we need to solve $2 = \frac{\sin(x)}{1-\sin(x)}$.

That becomes $2 - 2\sin(x)= \sin(x)$ so $\sin(x)=\frac{2}{3}$; this has two solutions in the given domain.

* Thanks to Andrew for pointing out a typo.