I had trouble remembering these too. Although it is certainly good to know how to work them out, their derivation is a little complicated and fiddly so for an exam it is best to have them memorized. (These are really the only integrals you need to remember though - the rest are obvious.)

Here are two methods to evaluate them

1) Make the substitution $t=\tan (x/2)$.

We now find that $\mathrm{dt}/\mathrm{dx}={\sec }^2(x/2)/2$ so $\mathrm{dx}/\mathrm{dt}=2{\cos }^2(x/2)$. So, for instance, the first integral would become:

$\int 1/\sin x\mathrm{dx}=\int 1/(2\sin (x/2)\cos (x/2))\times \mathrm{dx}/\mathrm{dt}\mathrm{dt}$

(using the double angle formula for sin)

$=\int 2{\cos }^2(x/2)/(2\sin (x/2)\cos (x/2))\mathrm{dt}$

$=\int 1/t\mathrm{dt}$

$=\ln t+C$

$=\ln \tan (x/2)+C$

Another method is to write the trigonometric functions as complex exponentials, and then write $u=e^{ix}$. I'm afraid this next bit is going to make no sense unless you know that:

$e^{ix}=\cos x+i\sin x$ (*)

If you haven't seen this before, stick to the first method for the time being.

Anyway, the second integral is:

$\int 1/\cos x\mathrm{dx}=\int 2/(e^{ix}+e^{-ix})\mathrm{dx}$

(by re-writing (*) as $\cos x=(e^{ix}+e^{-ix})/2$ and substituting)

Letting $u=e^{ix}$ our integral now becomes:

$\int 2/(u+1/u)\mathrm{dx}$

But $x=-i\ln u$, so $\mathrm{dx}/\mathrm{du}=-i/u$ and our integral is:

$\int (2/(u+1/u))\times -i/u=\int -2i/(u^2+1)\mathrm{du}=-2i\arctan (e^{ix})$

Then use the fact that $\arctan (x)=i/2\ln ((i+x)/(i-x))$. So our integral is:

$\ln ((i+e^{ix})/(i-e^{ix})+C$

Letting $t=x-\pi /2$ this is now (dropping the constant):

$\ln ((1+e^{it})/(1-e^{it}))=\ln ((e^{it/2}+e^{-it/2})/(e^{-it/2}-e^{it/2}))=\ln (2\cos (t/2)/(-2i\sin (t/2)))=\ln (i\cot (t/2))=\ln (i\cot (x/2-\pi /4))$

So the final answer is (hopefully)

$\ln \cot (\pi /4-x/2)+C$

(where a factor of $-i$ has been taken out - this simply changes the constant).

$\int \sec x\mathrm{dx}=\int (\sec x+\tan x)\sec x/(\sec x+\tan x)\mathrm{dx}$

and notice that the right hand side is of the form

$\int f\text{'}(x)/f(x)\mathrm{dx}$

so the answer comes out to be $\ln |\sec x+\tan x|+$ arbitrary constant of integration.

Similarly,

$\int \mathrm{cosec}x\mathrm{dx}=\int ({\mathrm{cosec}}^{2}x+\cot x\mathrm{cosec}x)/(\mathrm{cosec}x+\cot x)\mathrm{dx}=-\ln |\mathrm{cosec}x+\cot x|+$ constant.

Finally,

$\int \cot x\mathrm{dx}=\int (\cos x)/(\sin x)\mathrm{dx}=\ln \left|\sin x\right|+$ constant.

$\int \mathrm{sech}x\mathrm{dx}=-\ln |\mathrm{sech}x+\tanh x|+$ constant.

$\int \mathrm{cosech}x\mathrm{dx}=-\ln |\mathrm{cosech}x+\coth x|+$ constant

$\int \coth x\mathrm{dx}=\ln \left|\sinh x\right|+$constant