Question

are there any good tips for taking a calculus 2 exam? What is an easy way to remember how to do trig substitutions?

Answer 1

... do your homework.

Answer 2

dont do homework

Answer 3

Practice, practice, practice.

Answer 4

there are only three trig substitution you need if you see something in the form of:
x^2-a^2, x^2+a^2, or a^2-x^2.

if you see x^2-a^2, use sec(theta)=x/a
if you see x^2+a^2 use tan(theta)=x/a
if you see a^2-x^2 use sin(theta)=x/a

now there are other probably outside of this section different substitution of trig that you might have to use but if just if you see one of those forms you should be able to use what i told you above
though sometimes a substitution like this can be avoided
when you have something and this form
2x/(x^2-a^2) and you would like to integrate
then you just need a normal substitution (or let better phrase an algebraic substitution though a trig substitution would probably turn out okay in the end you are creating more work than there is)
sometimes things can get trickier
it really is practice and more practice you need

but anyways you might do trig substitution elsewhere
for example to integrate:
\[\int\limits_{}^{}\tan^{-1}(x) dx\]
I would let \[x=\tan(\theta)\]
\[dx=\sec^2(\theta) d \theta \]
so we have
\[\int\limits_{}^{}\tan^{-1}(\tan(\theta)) \sec^2(\theta) d \theta=\int\limits_{}^{} \theta \sec^2(\theta) d \theta\]
then i would do integration by parts
\[\theta \tan(\theta)-\int\limits_{}^{} \tan(\theta) d \theta=\theta \tan(\theta)-\int\limits_{}^{}\frac{\sin(\theta)}{\cos(\theta)} d \theta\]
then i would let u=cos(theta) => du=-sin(theta) d theta
\[\int\limits_{}^{} \tan^{-1}(x) dx=\int\limits_{}^{}\theta \sec^2(\theta) d \theta=\theta \tan(\theta)-\int\limits_{}^{}\frac{-du}{u}\]
\[=\theta \tan(\theta)+\ln|u|+C=\theta \tan(\theta)+\ln|\cos(\theta)|+C\]
which we need to back in terms of x using a right triangle

recall i let x=tan(theta)


\[=\tan^{-1}(x)+\ln|\frac{adjacent of \theta}{hyp}|+C\]
\[=\tan^{-1}(x)+\ln|\frac{1}{\sqrt{x^2+1}}|+C\]

Answer 5

oops i forgot reaplacing tan(theta)

Answer 6

\[=\tan^{-1}(x) \cdot x+\ln|\frac{1}{\sqrt{x^2+1}}|+C\]

Answer 7

someones writing themselves a novel ... i hope theres a vampire in it

Answer 8

lol